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LECTURE NOTES

CVEN 3525/3535

STUCTURAL ENGINEERING I

c©VICTOR E. SAOUMA

Spring 1999

Dept. of Civil Environmental and Architectural Engineering

University of Colorado, Boulder, CO 80309-0428

June 5, 2001

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Blank Page

Victor Saouma Structural Engineering

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PREFACE

Whereas there are numerous excellent textbooks covering Structural Analysis, or Structural Design, Ifelt that there was a need for a single reference which

• Provides a succinct, yet rigorous, coverage of Structural Engineering.

• Combines, as much as possible, Analysis with Design.

• Presents numerous, carefully selected, example problems.

in a properly type set document.As such, and given the reluctance of undergraduate students to go through extensive verbage in

order to capture a key concept, I have opted for an unusual format, one in which each key idea is clearlydistinguishable. In addition, such a format will hopefully foster group learning among students who caneasily reference misunderstood points.

Finally, whereas all problems have been taken from a variety of references, I have been very carefulin not only properly selecting them, but also in enhancing their solution through appropriate figures andLATEX typesetting macros.


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Structural Engineering can be characterized asthe art of molding materials we don’t really un-derstand into shapes we cannot really analyze soas to withstand forces we cannot really assess insuch a way that the public does not really sus-pect.

-Really Unknown Source


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Contents

I STRUCTURAL ENGINEERING I; CVEN 3525 0–3

1 INTRODUCTION 1–11.1 Structural Engineering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–11.2 Structures and their Surroundings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–11.3 Architecture & Engineering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–11.4 Architectural Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–21.5 Architectural Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–21.6 Structural Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–21.7 Structural Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–21.8 Load Transfer Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–31.9 Structure Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–31.10 Structural Engineering Courses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–101.11 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–12

2 LOADS 2–12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–12.2 Vertical Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–1

2.2.1 Dead Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–12.2.2 Live Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–2E 2-1 Live Load Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–42.2.3 Snow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–4

2.3 Lateral Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–62.3.1 Wind . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–6E 2-2 Wind Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–92.3.2 Earthquakes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–11E 2-3 Earthquake Load on a Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–14E 2-4 Earthquake Load on a Tall Building, (Schueller 1996) . . . . . . . . . . . . . . . . 2–15

2.4 Other Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–162.4.1 Hydrostatic and Earth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–16E 2-5 Hydrostatic Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–162.4.2 Thermal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–17E 2-6 Thermal Expansion/Stress (Schueller 1996) . . . . . . . . . . . . . . . . . . . . . . 2–172.4.3 Bridge Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–182.4.4 Impact Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–18

2.5 Other Important Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–182.5.1 Load Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–182.5.2 Load Placement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–192.5.3 Structural Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–192.5.4 Tributary Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–22

3 STRUCTURAL MATERIALS 3–13.1 Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–1

3.1.1 Structural Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–13.1.2 Reinforcing Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–2

3.2 Aluminum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–5

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3.3 Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–53.4 Masonry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–73.5 Timber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–73.6 Steel Section Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–7

3.6.1 ASCII File with Steel Section Properties . . . . . . . . . . . . . . . . . . . . . . . . 3–173.7 Joists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–19

4 EQUILIBRIUM & REACTIONS 4–14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–14.2 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–24.3 Equations of Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–34.4 Static Determinacy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–3

E 4-1 Statically Indeterminate Cable Structure . . . . . . . . . . . . . . . . . . . . . . . 4–34.5 Geometric Instability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–54.6 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–5

E 4-2 Simply Supported Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–5E 4-3 Parabolic Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–6E 4-4 Three Span Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–7E 4-5 Three Hinged Gable Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–8E 4-6 Inclined Supports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–10

4.7 Arches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–11

5 TRUSSES 5–15.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–1

5.1.1 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–15.1.2 Basic Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–1

5.2 Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–35.2.1 Determinacy and Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–35.2.2 Method of Joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–3E 5-1 Truss, Method of Joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–5

5.2.2.1 Matrix Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–7E 5-2 Truss I, Matrix Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–10E 5-3 Truss II, Matrix Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–115.2.3 Method of Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–13E 5-4 Truss, Method of Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–13

5.3 Case Study: Stadium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–14

6 CABLES 6–16.1 Funicular Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–1

E 6-1 Funicular Cable Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–16.2 Uniform Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–3

6.2.1 qdx; Parabola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–36.2.2 † qds; Catenary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–5

6.2.2.1 Historical Note . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–6E 6-2 Design of Suspension Bridge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–7

6.3 Case Study: George Washington Bridge . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–96.3.1 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–96.3.2 Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–106.3.3 Cable Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–106.3.4 Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–11

7 DESIGN PHILOSOPHIES of ACI and AISC CODES 7–17.1 Safety Provisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–17.2 Working Stress Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–27.3 Ultimate Strength Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–2

7.3.1 The Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–27.3.2 Reliability Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–47.3.3 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–5


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7.4 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–7E 7-1 LRFD vs ASD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–7

8 DESIGN I 8–18.1 Case Study: Eiffel Tower . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–1

8.1.1 Materials, & Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–18.1.2 Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–28.1.3 Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–58.1.4 Internal Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–68.1.5 Internal Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–7

8.2 Magazini Generali by Maillart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–88.2.1 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–88.2.2 Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–88.2.3 Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–98.2.4 Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–108.2.5 Internal Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–11

8.3 Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–128.3.1 Buildings Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–12

8.3.1.1 Wall Subsystems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–128.3.1.1.1 Example: Concrete Shear Wall . . . . . . . . . . . . . . . . . . . 8–138.3.1.1.2 Example: Trussed Shear Wall . . . . . . . . . . . . . . . . . . . 8–15

8.3.1.2 Shaft Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–158.3.1.2.1 Example: Tube Subsystem . . . . . . . . . . . . . . . . . . . . . 8–16

8.3.1.3 Rigid Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–178.4 Design of a Three Hinged Arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–178.5 Salginatobel Bridge (Maillart) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–18

8.5.1 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–188.5.2 Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–208.5.3 Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–218.5.4 Internal Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–218.5.5 Internal Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–248.5.6 Structural Behavior of Deck-Stiffened Arches . . . . . . . . . . . . . . . . . . . . . 8–25

9 STEEL TENSION MEMBERS 9–19.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–19.2 Geometric Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–1

9.2.1 Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–29.2.1.1 Net Area, An . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–2

E 9-1 Net Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–4E 9-2 Net Area, Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–5

9.2.1.2 Effective Net Area, Ae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–6E 9-3 Reduction Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–69.2.2 Stiffness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–6

9.3 LRFD Design of Tension Members . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–89.3.1 Tension Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–89.3.2 Block Shear Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–99.3.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–10E 9-4 Load Capacity of Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–10E 9-5 Shear Rupture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–12E 9-6 Shear Rupture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–13E 9-7 Complete Analysis/Design of a Truss . . . . . . . . . . . . . . . . . . . . . . . . . . 9–14

9.4 †† Computer Aided Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–20


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10 INTERNAL FORCES IN STRUCTURES 10–110.1 Design Sign Conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–110.2 Load, Shear, Moment Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–310.3 Moment Envelope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–410.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–4

10.4.1 Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–4E 10-1 Simple Shear and Moment Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . 10–4E 10-2 Sketches of Shear and Moment Diagrams . . . . . . . . . . . . . . . . . . . . . . . 10–610.4.2 Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–7E 10-3 Frame Shear and Moment Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–8E 10-4 Frame Shear and Moment Diagram; Hydrostatic Load . . . . . . . . . . . . . . . . 10–10E 10-5 Shear Moment Diagrams for Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–13E 10-6 Shear Moment Diagrams for Inclined Frame . . . . . . . . . . . . . . . . . . . . . . 10–1410.4.3 3D Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–15E 10-7 3D Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–15

10.5 Arches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–18

11 ARCHES and CURVED STRUCTURES 11–111.1 Arches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–1

11.1.1 Statically Determinate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–4E 11-1 Three Hinged Arch, Point Loads. (Gerstle 1974) . . . . . . . . . . . . . . . . . . . 11–4E 11-2 Semi-Circular Arch, (Gerstle 1974) . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–411.1.2 Statically Indeterminate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–6E 11-3 Statically Indeterminate Arch, (Kinney 1957) . . . . . . . . . . . . . . . . . . . . . 11–6

11.2 Curved Space Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–9E 11-4 Semi-Circular Box Girder, (Gerstle 1974) . . . . . . . . . . . . . . . . . . . . . . . 11–911.2.1 Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–10

11.2.1.1 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–1111.2.1.2 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–11

E 11-5 Internal Forces in an Helicoidal Cantilevered Girder, (Gerstle 1974) . . . . . . . . 11–12

12 DEFLECTION of STRUCTRES; Geometric Methods 12–112.1 Flexural Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–1

12.1.1 Curvature Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–112.1.2 Differential Equation of the Elastic Curve . . . . . . . . . . . . . . . . . . . . . . . 12–312.1.3 Moment Temperature Curvature Relation . . . . . . . . . . . . . . . . . . . . . . . 12–3

12.2 Flexural Deformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–412.2.1 Direct Integration Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–4E 12-1 Double Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–412.2.2 Curvature Area Method (Moment Area) . . . . . . . . . . . . . . . . . . . . . . . . 12–5

12.2.2.1 First Moment Area Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 12–512.2.2.2 Second Moment Area Theorem . . . . . . . . . . . . . . . . . . . . . . . . 12–5

E 12-2 Moment Area, Cantilevered Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–8E 12-3 Moment Area, Simply Supported Beam . . . . . . . . . . . . . . . . . . . . . . . . 12–8

12.2.2.3 Maximum Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–10E 12-4 Maximum Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–10E 12-5 Frame Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–11E 12-6 Frame Subjected to Temperature Loading . . . . . . . . . . . . . . . . . . . . . . . 12–1212.2.3 Elastic Weight/Conjugate Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–14E 12-7 Conjugate Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–15

12.3 Axial Deformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–1712.4 Torsional Deformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–17


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13 ENERGY METHODS; Part I 13–113.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–113.2 Real Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–1

13.2.1 External Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–113.2.2 Internal Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–2E 13-1 Deflection of a Cantilever Beam, (Chajes 1983) . . . . . . . . . . . . . . . . . . . . 13–3

13.3 Virtual Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–413.3.1 External Virtual Work δW

∗. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–6

13.3.2 Internal Virtual Work δU∗

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–613.3.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–9E 13-2 Beam Deflection (Chajes 1983) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–9E 13-3 Deflection of a Frame (Chajes 1983) . . . . . . . . . . . . . . . . . . . . . . . . . . 13–11E 13-4 Rotation of a Frame (Chajes 1983) . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–12E 13-5 Truss Deflection (Chajes 1983) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–13E 13-6 Torsional and Flexural Deformation, (Chajes 1983) . . . . . . . . . . . . . . . . . . 13–14E 13-7 Flexural and Shear Deformations in a Beam (White, Gergely and Sexmith 1976) . 13–15E 13-8 Thermal Effects in a Beam (White et al. 1976) . . . . . . . . . . . . . . . . . . . . 13–16E 13-9 Deflection of a Truss (White et al. 1976) . . . . . . . . . . . . . . . . . . . . . . . . 13–17E 13-10Thermal Defelction of a Truss; I (White et al. 1976) . . . . . . . . . . . . . . . . . 13–18E 13-11Thermal Deflections in a Truss; II (White et al. 1976) . . . . . . . . . . . . . . . . 13–19E 13-12Truss with initial camber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–20E 13-13Prestressed Concrete Beam with Continously Variable I (White et al. 1976) . . . . 13–21

13.4 *Maxwell Betti’s Reciprocal Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–2413.5 Summary of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–24

14 BRACED ROLLED STEEL BEAMS 14–114.1 Review from Strength of Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–1

14.1.1 Flexure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–114.1.2 Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–3

14.2 Nominal Strength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–414.3 Flexural Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–5

14.3.1 Failure Modes and Classification of Steel Beams . . . . . . . . . . . . . . . . . . . 14–514.3.1.1 Compact Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–514.3.1.2 Partially Compact Section . . . . . . . . . . . . . . . . . . . . . . . . . . 14–614.3.1.3 Slender Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–7

14.3.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–7E 14-1 Shape Factors, Rectangular Section . . . . . . . . . . . . . . . . . . . . . . . . . . 14–7E 14-2 Shape Factors, T Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–8E 14-3 Beam Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–9

14.4 Shear Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–1114.5 Deflections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–1114.6 Complete Design Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–12

15 COLUMN STABILITY 15–115.1 Introduction; Discrete Rigid Bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–1

15.1.1 Single Bar System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–115.1.2 Two Bars System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–215.1.3 ‡Analogy with Free Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–4

15.2 Continuous Linear Elastic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–515.2.1 Lower Order Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–615.2.2 Higher Order Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–6

15.2.2.1 Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–615.2.2.2 Hinged-Hinged Column . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–815.2.2.3 Fixed-Fixed Column . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–815.2.2.4 Fixed-Hinged Column . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–9

15.2.3 Effective Length Factors K . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–1015.3 Inelastic Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–12


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16 STEEL COMPRESSION MEMBERS 16–116.1 AISC Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16–116.2 LRFD Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16–216.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16–6

16.3.1 Evaluation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16–6E 16-1 Unbraced Column Evaluation, (?) . . . . . . . . . . . . . . . . . . . . . . . . . . . 16–6E 16-2 Braced Column Evaluation, (?) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16–6E 16-3 Braced Column Evaluation, (?) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16–716.3.2 Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16–8E 16-4 Column Design, (?) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16–8E 16-5 Design of Braced Column, (?) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16–9E 16-6 Design of a Column, (?) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16–9E 16-7 Column Design Using AISC Charts, (?) . . . . . . . . . . . . . . . . . . . . . . . . 16–10

17 STEEL CONNECTIONS 17–117.1 Bolted Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17–1

17.1.1 Types of Bolts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17–117.1.2 Types of Bolted Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17–117.1.3 Nominal Strength of Individual Bolts . . . . . . . . . . . . . . . . . . . . . . . . . . 17–417.1.4 Concentric Loads (Tension Connections) . . . . . . . . . . . . . . . . . . . . . . . . 17–5

17.1.4.1 AISC Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17–517.1.4.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17–6

E 17-1 Analysis of Bolted Connections (Salmon and Johnson 1990) . . . . . . . . . . . . . 17–6E 17-2 Design of Bolted Connections (Salmon and Johnson 1990) . . . . . . . . . . . . . . 17–817.1.5 Eccentric Connection (Shear Connections) . . . . . . . . . . . . . . . . . . . . . . . 17–9

17.2 Welded Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17–9

18 UNBRACED ROLLED STEEL BEAMS 18–118.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18–118.2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18–118.3 AISC Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18–2

18.3.1 Dividing values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18–218.3.2 Governing Moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18–2

18.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18–418.4.1 Verification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18–4E 18-1 Adequacy of an unbraced beam, (?) . . . . . . . . . . . . . . . . . . . . . . . . . . 18–4E 18-2 Adequacy of an unbraced beam, II (?) . . . . . . . . . . . . . . . . . . . . . . . . . 18–518.4.2 Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18–6E 18-3 Design of Laterally Unsupported Beam, (?) . . . . . . . . . . . . . . . . . . . . . . 18–7

18.5 Summary of AISC Governing Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18–11

19 Beam Columns, (Unedited) 19–119.1 Potential Modes of Failures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19–119.2 AISC Specifications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19–119.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19–1

19.3.1 Verification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19–1E 19-1 Verification, (?) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19–1E 19-2 8.2, (?) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19–419.3.2 Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19–6E 19-3 Design of Steel Beam-Column, (?) . . . . . . . . . . . . . . . . . . . . . . . . . . . 19–7

II STRUCTURAL ENGINEERING II; CVEN 3235 19–11

20 ARCHES and CURVED STRUCTURES 20–120.1 Arches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20–1

20.1.1 Statically Determinate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20–4E 20-1 Three Hinged Arch, Point Loads. (Gerstle 1974) . . . . . . . . . . . . . . . . . . . 20–4


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E 20-2 Semi-Circular Arch, (Gerstle 1974) . . . . . . . . . . . . . . . . . . . . . . . . . . . 20–420.1.2 Statically Indeterminate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20–6E 20-3 Statically Indeterminate Arch, (Kinney 1957) . . . . . . . . . . . . . . . . . . . . . 20–6

20.2 Curved Space Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20–9E 20-4 Semi-Circular Box Girder, (Gerstle 1974) . . . . . . . . . . . . . . . . . . . . . . . 20–920.2.1 Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20–10

20.2.1.1 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20–1120.2.1.2 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20–11

E 20-5 Internal Forces in an Helicoidal Cantilevered Girder, (Gerstle 1974) . . . . . . . . 20–12

21 STATIC INDETERMINANCY; FLEXIBILITY METHOD 21–121.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–121.2 The Force/Flexibility Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–421.3 Short-Cut for Displacement Evaluation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–521.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–5

E 21-1 Steel Building Frame Analysis, (White et al. 1976) . . . . . . . . . . . . . . . . . . 21–5E 21-2 Analysis of Irregular Building Frame, (White et al. 1976) . . . . . . . . . . . . . . 21–9E 21-3 Redundant Truss Analysis, (White et al. 1976) . . . . . . . . . . . . . . . . . . . . 21–11E 21-4 Truss with Two Redundants, (White et al. 1976) . . . . . . . . . . . . . . . . . . . 21–13E 21-5 Analysis of Nonprismatic Members, (White et al. 1976) . . . . . . . . . . . . . . . 21–16E 21-6 Fixed End Moments for Nonprismatic Beams, (White et al. 1976) . . . . . . . . . 21–17E 21-7 Rectangular Frame; External Load, (White et al. 1976) . . . . . . . . . . . . . . . 21–18E 21-8 Frame with Temperature Effects

and Support Displacements, (White et al. 1976) . . . . . . . . . . . . . . . . . . . 21–21E 21-9 Braced Bent with Loads and Temperature Change, (White et al. 1976) . . . . . . 21–23

22 APPROXIMATE FRAME ANALYSIS 22–122.1 Vertical Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22–122.2 Horizontal Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22–5

22.2.1 Portal Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22–8E 22-1 Approximate Analysis of a Frame subjected to Vertical and Horizontal Loads . . . 22–10

23 KINEMATIC INDETERMINANCY; STIFFNESS METHOD 23–123.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–1

23.1.1 Stiffness vs Flexibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–123.1.2 Sign Convention . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–2

23.2 Degrees of Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–223.2.1 Methods of Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–3

23.3 Kinematic Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–323.3.1 Force-Displacement Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–323.3.2 Fixed End Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–6

23.3.2.1 Uniformly Distributed Loads . . . . . . . . . . . . . . . . . . . . . . . . . 23–723.3.2.2 Concentrated Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–7

23.4 Slope Deflection; Direct Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–823.4.1 Slope Deflection Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–823.4.2 Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–823.4.3 Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–923.4.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–10E 23-1 Propped Cantilever Beam, (Arbabi 1991) . . . . . . . . . . . . . . . . . . . . . . . 23–10E 23-2 Two-Span Beam, Slope Deflection, (Arbabi 1991) . . . . . . . . . . . . . . . . . . . 23–10E 23-3 Two-Span Beam, Slope Deflection, Initial Deflection, (Arbabi 1991) . . . . . . . . 23–12E 23-4 dagger Frames, Slope Deflection, (Arbabi 1991) . . . . . . . . . . . . . . . . . . . . 23–13

23.5 Moment Distribution; Indirect Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–1523.5.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–15

23.5.1.1 Sign Convention . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–1523.5.1.2 Fixed-End Moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–1523.5.1.3 Stiffness Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–15


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23.5.1.4 Distribution Factor (DF) . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–1623.5.1.5 Carry-Over Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–16

23.5.2 Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–1723.5.3 Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–1723.5.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–18E 23-5 Continuous Beam, (Kinney 1957) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–18E 23-6 Continuous Beam, Simplified Method, (Kinney 1957) . . . . . . . . . . . . . . . . . 23–20E 23-7 Continuous Beam, Initial Settlement, (Kinney 1957) . . . . . . . . . . . . . . . . . 23–22E 23-8 Frame, (Kinney 1957) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–23E 23-9 Frame with Side Load, (Kinney 1957) . . . . . . . . . . . . . . . . . . . . . . . . . 23–27E 23-10Moment Distribution on a Spread-Sheet . . . . . . . . . . . . . . . . . . . . . . . . 23–29

24 REINFORCED CONCRETE BEAMS; Part I 24–124.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24–1

24.1.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24–124.1.2 Modes of Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24–224.1.3 Analysis vs Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24–224.1.4 Basic Relations and Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24–224.1.5 ACI Code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24–3

24.2 Cracked Section, Ultimate Strength Design Method . . . . . . . . . . . . . . . . . . . . . . 24–324.2.1 Equivalent Stress Block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24–324.2.2 Balanced Steel Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24–524.2.3 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24–524.2.4 Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24–6E 24-1 Ultimate Strength Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24–7E 24-2 Beam Design I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24–8E 24-3 Beam Design II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24–8

24.3 ACI Code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24–9

25 REINFORCED CONCRETE BEAMS; Part II 25–125.1 T Beams, (ACI 8.10) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25–1

25.1.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25–325.1.2 Design, (balanced) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25–3E 25-1 T Beam; Moment Capacity I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25–3E 25-2 T Beam; Moment Capacity II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25–4E 25-3 T Beam; Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25–5

25.2 Doubly Reinforced Rectangular Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25–625.2.1 Tests for fs and f ′s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25–725.2.2 Moment Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25–9E 25-4 Doubly Reinforced Concrete beam; Review . . . . . . . . . . . . . . . . . . . . . . 25–10E 25-5 Doubly Reinforced Concrete beam; Design . . . . . . . . . . . . . . . . . . . . . . . 25–11

26 DIRECT STIFFNESS METHOD 26–126.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–1

26.1.1 Structural Idealization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–126.1.2 Structural Discretization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–226.1.3 Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–226.1.4 Sign Convention . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–326.1.5 Degrees of Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–3

26.2 Stiffness Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–526.2.1 Truss Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–526.2.2 Beam Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–726.2.3 2D Frame Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–826.2.4 Remarks on Element Stiffness Matrices . . . . . . . . . . . . . . . . . . . . . . . . 26–8

26.3 Direct Stiffness Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–826.3.1 Orthogonal Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–8E 26-1 Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–10


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26.3.2 Local and Global Element Stiffness Matrices ([k(e)] [K(e)]) . . . . . . . . . . . . . 26–1226.3.2.1 2D Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–12

26.3.3 Global Stiffness Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–1326.3.3.1 Structural Stiffness Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . 26–1326.3.3.2 Augmented Stiffness Matrix . . . . . . . . . . . . . . . . . . . . . . . . . 26–13

26.3.4 Internal Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–1426.3.5 Boundary Conditions, [ID] Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–1526.3.6 LM Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–1626.3.7 Assembly of Global Stiffness Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . 26–16E 26-2 Assembly of the Global Stiffness Matrix . . . . . . . . . . . . . . . . . . . . . . . . 26–1726.3.8 Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–18E 26-3 Direct Stiffness Analysis of a Truss . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–18E 26-4 Analysis of a Frame with MATLAB . . . . . . . . . . . . . . . . . . . . . . . . . . 26–23E 26-5 Analysis of a simple Beam with Initial Displacements . . . . . . . . . . . . . . . . 26–25

26.4 Computer Program Organization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–2926.5 Computer Implementation with MATLAB . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–31

26.5.1 Program Input . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–3126.5.1.1 Input Variable Descriptions . . . . . . . . . . . . . . . . . . . . . . . . . . 26–3126.5.1.2 Sample Input Data File . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–3226.5.1.3 Program Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–33

26.5.2 Program Listing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–3426.5.2.1 Main Program . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–3426.5.2.2 Assembly of ID Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–3526.5.2.3 Element Nodal Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . 26–3626.5.2.4 Element Lengths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–3726.5.2.5 Element Stiffness Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . 26–3726.5.2.6 Transformation Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–3826.5.2.7 Assembly of the Augmented Stiffness Matrix . . . . . . . . . . . . . . . . 26–3926.5.2.8 Print General Information . . . . . . . . . . . . . . . . . . . . . . . . . . 26–4026.5.2.9 Print Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–4026.5.2.10 Load Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–4126.5.2.11 Nodal Displacements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–4226.5.2.12 Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–4326.5.2.13 Internal Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–4426.5.2.14 Plotting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–4526.5.2.15 Sample Output File . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–49

27 COLUMNS 27–127.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–1

27.1.1 Types of Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–127.1.2 Possible Arrangement of Bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–1

27.2 Short Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–327.2.1 Concentric Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–327.2.2 Eccentric Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–3

27.2.2.1 Balanced Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–427.2.2.2 Tension Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–627.2.2.3 Compression Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–7

27.2.3 ACI Provisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–727.2.4 Interaction Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–827.2.5 Design Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–8E 27-1 R/C Column, c known . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–8E 27-2 R/C Column, e known . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–9E 27-3 R/C Column, Using Design Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–14


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28 DESIGN II 28–128.1 Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28–1

28.1.1 Beam Column Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28–128.1.2 Behavior of Simple Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28–128.1.3 Eccentricity of Applied Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28–228.1.4 Design of a Statically Indeterminate Arch . . . . . . . . . . . . . . . . . . . . . . . 28–428.1.5 Temperature Changes in an Arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28–10

29 INFLUENCE LINES (unedited) 29–1

30 ELEMENTS of STRUCTURAL RELIABILITY 30–130.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–130.2 Elements of Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–130.3 Distributions of Random Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–3

30.3.1 Uniform Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–330.3.2 Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–330.3.3 Lognormal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–430.3.4 Beta Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–430.3.5 BiNormal distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–4

30.4 Reliability Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–430.4.1 Performance Function Identification . . . . . . . . . . . . . . . . . . . . . . . . . . 30–430.4.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–530.4.3 Mean and Standard Deviation of a Performance Function . . . . . . . . . . . . . . 30–6

30.4.3.1 Direct Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–630.4.3.2 Monte Carlo Simulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–630.4.3.3 Point Estimate Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–930.4.3.4 Taylor’s Series-Finite Difference Estimation . . . . . . . . . . . . . . . . . 30–9

30.4.4 Overall System Reliability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–1030.4.5 Target Reliability Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–10

30.5 Reliability Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–12


Draft

List of Figures

1.1 Types of Forces in Structural Elements (1D) . . . . . . . . . . . . . . . . . . . . . . . . . . 1–31.2 Basic Aspects of Cable Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–41.3 Basic Aspects of Arches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–51.4 Types of Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–61.5 Variations in Post and Beams Configurations . . . . . . . . . . . . . . . . . . . . . . . . . 1–71.6 Different Beam Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–81.7 Basic Forms of Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–91.8 Examples of Air Supported Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–101.9 Basic Forms of Shells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–111.10 Sequence of Structural Engineering Courses . . . . . . . . . . . . . . . . . . . . . . . . . . 1–11

2.1 Approximation of a Series of Closely Spaced Loads . . . . . . . . . . . . . . . . . . . . . . 2–22.2 Snow Map of the United States, ubc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–42.3 Loads on Projected Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–52.4 Vertical and Normal Loads Acting on Inclined Surfaces . . . . . . . . . . . . . . . . . . . 2–52.5 Wind Map of the United States, (UBC 1995) . . . . . . . . . . . . . . . . . . . . . . . . . 2–62.6 Effect of Wind Load on Structures(Schueller 1996) . . . . . . . . . . . . . . . . . . . . . . 2–72.7 Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Struc-

tures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–102.8 Vibrations of a Building . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–112.9 Seismic Zones of the United States, (UBC 1995) . . . . . . . . . . . . . . . . . . . . . . . 2–122.10 Earth and Hydrostatic Loads on Structures . . . . . . . . . . . . . . . . . . . . . . . . . . 2–172.11 Truck Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–182.12 Load Placement to Maximize Moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–202.13 Load Life of a Structure, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . 2–202.14 Concept of Tributary Areas for Structural Member Loading . . . . . . . . . . . . . . . . . 2–212.15 One or Two Way actions in Slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–212.16 Load Transfer in R/C Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–232.17 Two Way Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–242.18 Example of Load Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–24

3.1 Stress Strain Curves of Concrete and Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–13.2 Standard Rolled Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–23.3 Residual Stresses in Rolled Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–33.4 Residual Stresses in Welded Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–33.5 Influence of Residual Stress on Average Stress-Strain Curve of a Rolled Section . . . . . . 3–43.6 Concrete Stress-Strain curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–63.7 Concrete microcracking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–63.8 W and C sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–83.9 prefabricated Steel Joists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–20

4.1 Types of Supports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–14.2 Inclined Roller Support . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–34.3 Examples of Static Determinate and Indeterminate Structures . . . . . . . . . . . . . . . . 4–44.4 Geometric Instability Caused by Concurrent Reactions . . . . . . . . . . . . . . . . . . . . 4–5

Draft0–2 LIST OF FIGURES

5.1 Types of Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–25.2 Bridge Truss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–25.3 A Statically Indeterminate Truss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–45.4 X and Y Components of Truss Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–45.5 Sign Convention for Truss Element Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–55.6 Direction Cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–85.7 Forces Acting on Truss Joint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–85.8 Complex Statically Determinate Truss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–95.9 Florence Stadium, Pier Luigi Nervi (?) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–145.10 Florence Stadioum, Pier Luigi Nervi (?) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–15

6.1 Cable Structure Subjected to q(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–36.2 Catenary versus Parabola Cable Structures . . . . . . . . . . . . . . . . . . . . . . . . . . 6–66.3 Leipniz’s Figure of a catenary, 1690 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–76.4 Longitudinal and Plan Elevation of the George Washington Bridge . . . . . . . . . . . . . 6–96.5 Truck Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–106.6 Dead and Live Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–116.7 Location of Cable Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–116.8 Vertical Reactions in Columns Due to Central Span Load . . . . . . . . . . . . . . . . . . 6–126.9 Cable Reactions in Side Span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–136.10 Cable Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–136.11 Deck Idealization, Shear and Moment Diagrams . . . . . . . . . . . . . . . . . . . . . . . . 6–14

7.1 Load Life of a Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–27.2 Normalized Gauss Distribution, and Cumulative Distribution Function . . . . . . . . . . . 7–37.3 Frequency Distributions of Load Q and Resistance R . . . . . . . . . . . . . . . . . . . . . 7–47.4 Definition of Reliability Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–57.5 Probability of Failure in terms of β . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–5

8.1 Eiffel Tower (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–18.2 Eiffel Tower Idealization, (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . . . 8–38.3 Eiffel Tower, Dead Load Idealization; (Billington and Mark 1983) . . . . . . . . . . . . . . 8–38.4 Eiffel Tower, Wind Load Idealization; (Billington and Mark 1983) . . . . . . . . . . . . . . 8–48.5 Eiffel Tower, Wind Loads, (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . . 8–48.6 Eiffel Tower, Reactions; (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . . . 8–58.7 Eiffel Tower, Internal Gravity Forces; (Billington and Mark 1983) . . . . . . . . . . . . . . 8–68.8 Eiffel Tower, Horizontal Reactions; (Billington and Mark 1983) . . . . . . . . . . . . . . . 8–68.9 Eiffel Tower, Internal Wind Forces; (Billington and Mark 1983) . . . . . . . . . . . . . . . 8–78.10 Magazzini Generali; Overall Dimensions, (Billington and Mark 1983) . . . . . . . . . . . . 8–88.11 Magazzini Generali; Support System, (Billington and Mark 1983) . . . . . . . . . . . . . . 8–98.12 Magazzini Generali; Loads (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . . 8–98.13 Magazzini Generali; Beam Reactions, (Billington and Mark 1983) . . . . . . . . . . . . . . 8–98.14 Magazzini Generali; Shear and Moment Diagrams (Billington and Mark 1983) . . . . . . . 8–108.15 Magazzini Generali; Internal Moment, (Billington and Mark 1983) . . . . . . . . . . . . . 8–108.16 Magazzini Generali; Similarities Between The Frame Shape and its Moment Diagram,

(Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–118.17 Magazzini Generali; Equilibrium of Forces at the Beam Support, (Billington and Mark

1983) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–118.18 Magazzini Generali; Effect of Lateral Supports, (Billington and Mark 1983) . . . . . . . . 8–128.19 Design of a Shear Wall Subsystem, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . 8–138.20 Trussed Shear Wall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–158.21 Design Example of a Tubular Structure, (Lin and Stotesbury 1981) . . . . . . . . . . . . . 8–168.22 A Basic Portal Frame, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . . . 8–178.23 Salginatobel Bridge; Dimensions, (Billington and Mark 1983) . . . . . . . . . . . . . . . . 8–198.24 Salginatobel Bridge; Idealization, (Billington and Mark 1983) . . . . . . . . . . . . . . . . 8–198.25 Salginatobel Bridge; Hinges, (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . 8–198.26 Salginatobel Bridge; Sections, (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . 8–20


DraftLIST OF FIGURES 0–3

8.27 Salginatobel Bridge; Dead Load, (Billington and Mark 1983) . . . . . . . . . . . . . . . . 8–208.28 Salginatobel Bridge; Truck Load, (Billington and Mark 1983) . . . . . . . . . . . . . . . . 8–228.29 Salginatobel Bridge; Total Vertical Load, (Billington and Mark 1983) . . . . . . . . . . . . 8–238.30 Salginatobel Bridge; Reactions, (Billington and Mark 1983) . . . . . . . . . . . . . . . . . 8–238.31 Salganitobel Bridge; Shear Diagrams, (Billington and Mark 1983) . . . . . . . . . . . . . . 8–238.32 Salginatobel Bridge; Live Load Moment Diagram, (Billington and Mark 1983) . . . . . . . 8–248.33 Structural Behavior of Stiffened Arches, (Billington 1979) . . . . . . . . . . . . . . . . . . 8–25

9.1 Stress Concentration Around Circular Hole . . . . . . . . . . . . . . . . . . . . . . . . . . 9–29.2 Hole Sizes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–39.3 Effect of Staggered Holes on Net Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–39.4 Gage Distances for an Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–49.5 Net and Gross Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–99.6 Tearing Failure Limit State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–9

10.1 Shear and Moment Sign Conventions for Design . . . . . . . . . . . . . . . . . . . . . . . . 10–210.2 Sign Conventions for 3D Frame Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–210.3 Free Body Diagram of an Infinitesimal Beam Segment . . . . . . . . . . . . . . . . . . . . 10–310.4 Shear and Moment Forces at Different Sections of a Loaded Beam . . . . . . . . . . . . . 10–410.5 Slope Relations Between Load Intensity and Shear, or Between Shear and Moment . . . . 10–510.6 Inclined Loads on Inclined Members . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–8

11.1 Moment Resisting Forces in an Arch or Suspension System as Compared to a Beam, (Linand Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–2

11.2 Statics of a Three-Hinged Arch, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . 11–211.3 Two Hinged Arch, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . . . . . 11–311.4 Arch Rib Stiffened with Girder or Truss, (Lin and Stotesbury 1981) . . . . . . . . . . . . 11–311.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–411.6 Semi-Circular three hinged arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–511.7 Semi-Circular three hinged arch; Free body diagram . . . . . . . . . . . . . . . . . . . . . 11–611.8 Statically Indeterminate Arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–711.9 Statically Indeterminate Arch; ‘Horizontal Reaction Removed . . . . . . . . . . . . . . . . 11–811.10Semi-Circular Box Girder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–911.11Geometry of Curved Structure in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–1111.12Free Body Diagram of a Curved Structure in Space . . . . . . . . . . . . . . . . . . . . . . 11–1211.13Helicoidal Cantilevered Girder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–13

12.1 Curvature of a flexural element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–212.2 Moment Area Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–612.3 Sign Convention for the Moment Area Method . . . . . . . . . . . . . . . . . . . . . . . . 12–712.4 Areas and Centroid of Polynomial Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–712.5 Maximum Deflection Using the Moment Area Method . . . . . . . . . . . . . . . . . . . . 12–1012.6 Conjugate Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–1512.7 Torsion Rotation Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–17

13.1 Load Deflection Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–213.2 Strain Energy Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–313.3 Deflection of Cantilever Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–413.4 Real and Virtual Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–513.5 Torsion Rotation Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–713.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–1013.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–1013.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–1113.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–1213.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–1313.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–1413.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–1513.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–17



13.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–1813.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–1913.16*(correct 42.7 to 47.2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–22

14.1 Bending of a Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–214.2 Stress distribution at different stages of loading . . . . . . . . . . . . . . . . . . . . . . . . 14–214.3 Stress-strain diagram for most structural steels . . . . . . . . . . . . . . . . . . . . . . . . 14–314.4 Flexural and Shear Stress Distribution in a Rectangular Beam . . . . . . . . . . . . . . . 14–414.5 Local (flange) Buckling; Flexural and Torsional Buckling Modes in a Rolled Section,

(Lulea University) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–514.6 W Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–614.7 Nominal Moments for Compact and Partially Compact Sections . . . . . . . . . . . . . . . 14–714.8 AISC Requirements for Shear Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–12

15.1 Stability of a Rigid Bar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–115.2 Stability of a Rigid Bar with Initial Imperfection . . . . . . . . . . . . . . . . . . . . . . . 15–215.3 Stability of a Two Rigid Bars System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–315.4 Two DOF Dynamic System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–415.5 Euler Column . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–515.6 Simply Supported Beam Column; Differential Segment; Effect of Axial Force P . . . . . . 15–715.7 Solution of the Tanscendental Equation for the Buckling Load of a Fixed-Hinged Column 15–1015.8 Column Effective Lengths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–1115.9 Frame Effective Lengths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–1115.10Column Effective Length in a Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–1215.11Standard Alignment Chart (AISC) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–1315.12Inelastic Buckling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–1315.13Euler Buckling, and SSRC Column Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–14

16.1 SSRC Column Curve and AISC Critical Stresses . . . . . . . . . . . . . . . . . . . . . . . 16–216.2 Fcr versus KL/r According to LRFD, for Various Fy . . . . . . . . . . . . . . . . . . . . . 16–5

17.1 Examples of Bolted Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17–217.2 Stress Transfer by Shear and Bearing in a Bolted Connection . . . . . . . . . . . . . . . . 17–217.3 Stress Transfer in a Friction Type Bolted Connection . . . . . . . . . . . . . . . . . . . . . 17–317.4 Modes of Failure of Bolted Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17–317.5 Number of Shearing Planes m in Bolted Connections . . . . . . . . . . . . . . . . . . . . . 17–417.6 Bearing Strength Related to End Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . 17–417.7 Basic Types of Weld . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17–9

18.1 I Shaped Beam in Slightly Buckled Position . . . . . . . . . . . . . . . . . . . . . . . . . . 18–118.2 Nominal Strength Mn of “Compact” Sections as Affected by Lateral-Torsional Buckling . 18–318.3 Design of Laterally Unsupported Steel Beam . . . . . . . . . . . . . . . . . . . . . . . . . 18–7

19.1 Rigid Frame Subjected to Lateral Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . 19–4

20.1 Moment Resisting Forces in an Arch or Suspension System as Compared to a Beam, (Linand Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20–2

20.2 Statics of a Three-Hinged Arch, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . 20–220.3 Two Hinged Arch, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . . . . . 20–320.4 Arch Rib Stiffened with Girder or Truss, (Lin and Stotesbury 1981) . . . . . . . . . . . . 20–320.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20–420.6 Semi-Circular three hinged arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20–520.7 Semi-Circular three hinged arch; Free body diagram . . . . . . . . . . . . . . . . . . . . . 20–620.8 Statically Indeterminate Arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20–720.9 Statically Indeterminate Arch; ‘Horizontal Reaction Removed . . . . . . . . . . . . . . . . 20–820.10Semi-Circular Box Girder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20–920.11Geometry of Curved Structure in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20–1120.12Free Body Diagram of a Curved Structure in Space . . . . . . . . . . . . . . . . . . . . . . 20–12


DraftLIST OF FIGURES 0–5

20.13Helicoidal Cantilevered Girder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20–13

21.1 Statically Indeterminate 3 Cable Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–221.2 Propped Cantilever Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–321.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–621.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–621.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–721.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–821.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–921.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–1021.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–1221.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–1321.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–1421.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–1621.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–1621.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–1821.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–1921.16Definition of Flexibility Terms for a Rigid Frame . . . . . . . . . . . . . . . . . . . . . . . 21–1921.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–2221.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–23

22.1 Uniformly Loaded Beam and Frame with Free or Fixed Beam Restraint . . . . . . . . . . 22–222.2 Uniformly Loaded Frame, Approximate Location of Inflection Points . . . . . . . . . . . . 22–322.3 Approximate Analysis of Frames Subjected to Vertical Loads; Girder Moments . . . . . . 22–422.4 Approximate Analysis of Frames Subjected to Vertical Loads; Column Axial Forces . . . . 22–522.5 Approximate Analysis of Frames Subjected to Vertical Loads; Column Moments . . . . . 22–622.6 Horizontal Force Acting on a Frame, Approximate Location of Inflection Points . . . . . . 22–722.7 Approximate Analysis of Frames Subjected to Lateral Loads; Column Shear . . . . . . . . 22–822.8 ***Approximate Analysis of Frames Subjected to Lateral Loads; Girder Moment . . . . . 22–922.9 Approximate Analysis of Frames Subjected to Lateral Loads; Column Axial Force . . . . 22–1022.10Example; Approximate Analysis of a Building . . . . . . . . . . . . . . . . . . . . . . . . . 22–1022.11Free Body Diagram for the Approximate Analysis of a Frame Subjected to Vertical Loads 22–1122.12Approximate Analysis of a Building; Moments Due to Vertical Loads . . . . . . . . . . . . 22–1322.13Approximate Analysis of a Building; Shears Due to Vertical Loads . . . . . . . . . . . . . 22–1522.14Approximate Analysis for Vertical Loads; Spread-Sheet Format . . . . . . . . . . . . . . . 22–1622.15Approximate Analysis for Vertical Loads; Equations in Spread-Sheet . . . . . . . . . . . . 22–1722.16Free Body Diagram for the Approximate Analysis of a Frame Subjected to Lateral Loads 22–1822.17Approximate Analysis of a Building; Moments Due to Lateral Loads . . . . . . . . . . . . 22–2022.18Portal Method; Spread-Sheet Format . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22–2222.19Portal Method; Equations in Spread-Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . 22–23

23.1 Sign Convention, Design and Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–223.2 Independent Displacements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–223.3 Total Degrees of Freedom for various Type of Elements . . . . . . . . . . . . . . . . . . . 23–423.4 Flexural Problem Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–523.5 Illustrative Example for the Slope Deflection Method . . . . . . . . . . . . . . . . . . . . . 23–923.6 Slope Deflection; Propped Cantilever Beam . . . . . . . . . . . . . . . . . . . . . . . . . . 23–1023.7 Two-Span Beam, Slope Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–1123.8 Two Span Beam, Slope Deflection, Moment Diagram . . . . . . . . . . . . . . . . . . . . . 23–1223.9 Frame Analysis by the Slope Deflection Method . . . . . . . . . . . . . . . . . . . . . . . . 23–13

24.1 Cracked Section, Limit State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24–424.2 Whitney Stress Block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24–4

25.1 T Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25–125.2 T Beam as Rectangular Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25–225.3 T Beam Strain and Stress Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25–225.4 Decomposition of Steel Reinforcement for T Beams . . . . . . . . . . . . . . . . . . . . . . 25–3



25.5 Doubly Reinforced Beams; Strain and Stress Diagrams . . . . . . . . . . . . . . . . . . . . 25–725.6 Different Possibilities for Doubly Reinforced Concrete Beams . . . . . . . . . . . . . . . . 25–725.7 Strain Diagram, Doubly Reinforced Beam; is A′

s Yielding? . . . . . . . . . . . . . . . . . . 25–825.8 Summary of Conditions for top and Bottom Steel Yielding . . . . . . . . . . . . . . . . . . 25–9

26.1 Global Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–326.2 Local Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–426.3 Sign Convention, Design and Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–426.4 Total Degrees of Freedom for various Type of Elements . . . . . . . . . . . . . . . . . . . 26–426.5 Dependent Displacements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–526.6 Examples of Active Global Degrees of Freedom . . . . . . . . . . . . . . . . . . . . . . . . 26–626.7 Problem with 2 Global d.o.f. θ1 and θ2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–926.8 2D Frame Element Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–1326.9 *Frame Example (correct K32 and K33) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–1426.10Example for [ID] Matrix Determination . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–1626.11Simple Frame Anlysed with the MATLAB Code . . . . . . . . . . . . . . . . . . . . . . . 26–1726.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–1926.13Simple Frame Anlysed with the MATLAB Code . . . . . . . . . . . . . . . . . . . . . . . 26–2326.14ID Values for Simple Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–2626.15Structure Plotted with CASAP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–49

27.1 Types of columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–127.2 Tied vs Spiral Reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–227.3 Possible Bar arrangements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–227.4 Sources of Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–327.5 Load moment interaction diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–427.6 Strain and Stress Diagram of a R/C Column . . . . . . . . . . . . . . . . . . . . . . . . . 27–527.7 Column Interaction Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27–8

28.1 Flexible, Rigid, and Semi-Flexible Joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28–128.2 Deformation of Flexible and Rigid Frames Subjected to Vertical and Horizontal Loads,

(Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28–228.3 Deformation, Shear, Moment, and Axial Diagrams for Various Types of Portal Frames

Subjected to Vertical and Horizontal Loads . . . . . . . . . . . . . . . . . . . . . . . . . . 28–328.4 Axial and Flexural Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28–428.5 Design of a Statically Indeterminate Arch . . . . . . . . . . . . . . . . . . . . . . . . . . . 28–528.6 Normal and Shear Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28–7

30.1 Normalized Gauss Distribution, and Cumulative Distribution Function . . . . . . . . . . . 30–330.2 Definition of Reliability Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–530.3 Probability of Failure in terms of β . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–11


Draft

List of Tables

2.1 Unit Weight of Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–22.2 Weights of Building Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–32.3 Average Gross Dead Load in Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–32.4 Minimum Uniformly Distributed Live Loads, (UBC 1995) . . . . . . . . . . . . . . . . . . 2–32.5 Wind Velocity Variation above Ground . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–72.6 Ce Coefficients for Wind Load, (UBC 1995) . . . . . . . . . . . . . . . . . . . . . . . . . . 2–82.7 Wind Pressure Coefficients Cq, (UBC 1995) . . . . . . . . . . . . . . . . . . . . . . . . . . 2–82.8 Importance Factors for Wind and Earthquake Load, (UBC 1995) . . . . . . . . . . . . . . 2–82.9 Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Struc-

tures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–92.10 Z Factors for Different Seismic Zones, ubc . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–122.11 S Site Coefficients for Earthquake Loading, (UBC 1995) . . . . . . . . . . . . . . . . . . . 2–132.12 Partial List of RW for Various Structure Systems, (UBC 1995) . . . . . . . . . . . . . . . 2–132.13 Coefficients of Thermal Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–17

3.1 Properties of Major Structural Steels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–33.2 Properties of Reinforcing Bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–43.3 Joist Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–21

4.1 Equations of Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–2

5.1 Static Determinacy and Stability of Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . 5–3

7.1 Allowable Stresses for Steel and Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–37.2 Selected β values for Steel and Concrete Structures . . . . . . . . . . . . . . . . . . . . . . 7–67.3 Strength Reduction Factors, Φ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–6

9.1 Usual Gage Lengths g, g1, g2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–59.2 Effective Net Area Ae for Bolted and Welded Connections . . . . . . . . . . . . . . . . . . 9–7

12.1 Conjugate Beam Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–15

13.1 Possible Combinations of Real and Hypothetical Formulations . . . . . . . . . . . . . . . . 13–513.2 k Factors for Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–913.3 Summary of Expressions for the Internal Strain Energy and External Work . . . . . . . . 13–25

15.1 Essential and Natural Boundary Conditions for Columns . . . . . . . . . . . . . . . . . . . 15–8

16.1 Design Stress φFcr for Fy=36 ksi in Terms of KLrmin

. . . . . . . . . . . . . . . . . . . . . . 16–316.2 Design Stress φFcr for Fy=50 ksi in Terms of KL

rmin. . . . . . . . . . . . . . . . . . . . . . 16–4

16.3 Stiffness Reduction Factors (AISC) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16–5

17.1 Nominal Areas of Standard Bolts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17–1

18.1 Unbraced Beam Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18–918.2 Summary of AISC Governing Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18–11

Draft0–2 LIST OF TABLES

19.1 Values ofm for use in Beam-Column Design Equation, (from a paper in AISC EngineeringJournal by Uang, Wattar, and Leet (1990)) . . . . . . . . . . . . . . . . . . . . . . . . . . 19–7

21.1 Table of∫ L

0

g1(x)g2(x)dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–5

21.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–1521.3 Displacement Computations for a Rectangular Frame . . . . . . . . . . . . . . . . . . . . 21–20

22.1 Columns Combined Approximate Vertical and Horizontal Loads . . . . . . . . . . . . . . 22–2122.2 Girders Combined Approximate Vertical and Horizontal Loads . . . . . . . . . . . . . . . 22–24

23.1 Stiffness vs Flexibility Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–123.2 Degrees of Freedom of Different Structure Types Systems . . . . . . . . . . . . . . . . . . 23–3

26.1 Example of Nodal Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–226.2 Example of Element Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–226.3 Example of Group Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26–226.4 Degrees of Freedom of Different Structure Types Systems . . . . . . . . . . . . . . . . . . 26–7

28.1 Geometry of the Arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28–528.2 Calculation of Horizontal Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28–628.3 Moment at the Centers of the Ribs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28–728.4 Values of Normal and Shear Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28–828.5 Design of the Arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28–928.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28–1028.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28–1028.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28–1128.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28–11

30.1 Tabulated Results of a Simple Monte-Carlo Simulation. . . . . . . . . . . . . . . . . . . . 30–930.2 Selected β values for Steel and Concrete Structures . . . . . . . . . . . . . . . . . . . . . . 30–12


Draft

Part I

STRUCTURAL ENGINEERING I;CVEN 3525

Draft

Chapter 1

INTRODUCTION

1.1 Structural Engineering

1 Structural engineers are responsible for the detailed analysis and design of:

Architectural structures: Buildings, houses, factories. They must work in close cooperation with anarchitect who will ultimately be responsible for the design.

Civil Infrastructures: Bridges, dams, pipelines, offshore structures. They work with transportation,hydraulic, nuclear and other engineers. For those structures they play the leading role.

Aerospace, Mechanical, Naval structures: aeroplanes, spacecrafts, cars, ships, submarines to en-sure the structural safety of those important structures.

1.2 Structures and their Surroundings

2 Structural design is affected by various environmental constraints:

1. Major movements: For example, elevator shafts are usually shear walls good at resisting lateralload (wind, earthquake).

2. Sound and structure interact:

• A dome roof will concentrate the sound

• A dish roof will diffuse the sound

3. Natural light:

• A flat roof in a building may not provide adequate light.

• A Folded plate will provide adequate lighting (analysis more complex).

• A bearing and shear wall building may not have enough openings for daylight.

• A Frame design will allow more light in (analysis more complex).

4. Conduits for cables (electric, telephone, computer), HVAC ducts, may dictate type of floor system.

5. Net clearance between columns (unobstructed surface) will dictate type of framing.

1.3 Architecture & Engineering

3 Architecture must be the product of a creative collaboration of architects and engineers.

4 Architect stress the overall, rather than elemental approach to design. In the design process, theyconceptualize a space-form scheme as a total system. They are generalists.

Draft1–2 INTRODUCTION

5 The engineer, partly due to his/her education think in reverse, starting with details and withoutsufficient regards for the overall picture. (S)he is a pragmatist who “knows everything about nothing”.

6 Thus there is a conceptual gap between architects and engineers at all levels of design.

7 Engineer’s education is more specialized and in depth than the architect’s. However, engineer mustbe kept aware of overall architectural objective.

8 In the last resort, it is the architect who is the leader of the construction team, and the engineers arehis/her servant.

9 A possible compromise might be an Architectural Engineer.

1.4 Architectural Design Process

10 Architectural design is hierarchical:

Schematic: conceptual overall space-form feasibility of basic schematic options. Collaboration is mostlybetween the owner and the architect.

Preliminary: Establish basic physical properties of major subsystems and key components to provedesign feasibility. Some collaboration with engineers is necessary.

Final design: final in-depth design refinements of all subsystems and components and preparation ofworking documents (“blue-prints”). Engineers play a leading role.

1.5 Architectural Design

11 Architectural design must respect various constraints:

Functionality: Influence of the adopted structure on the purposes for which the structure was erected.

Aesthetics: The architect often imposes his aesthetic concerns on the engineer. This in turn can placesevere limitations on the structural system.

Economy: It should be kept in mind that the two largest components of a structure are labors andmaterials. Design cost is comparatively negligible.

1.6 Structural Analysis

12 Given an existing structure subjected to a certain load determine internal forces (axial, shear, flex-ural, torsional; or stresses), deflections, and verify that no unstable failure can occur.

13 Thus the basic structural requirements are:

Strength: stresses should not exceed critical values: σ < σf

Stiffness: deflections should be controlled: ∆ < ∆max

Stability: buckling or cracking should also be prevented

1.7 Structural Design

14 Given a set of forces, dimension the structural element.

Steel/wood Structures Select appropriate section.

Reinforced Concrete: Determine dimensions of the element and internal reinforcement (number andsizes of reinforcing bars).


Draft1.8 Load Transfer Elements 1–3

15 For new structures, iterative process between analysis and design. A preliminary design is madeusing rules of thumbs (best known to Engineers with design experience) and analyzed. Followingdesign, we check for

Serviceability: deflections, crack widths under the applied load. Compare with acceptable valuesspecified in the design code.

Failure: and compare the failure load with the applied load times the appropriate factors of safety.

If the design is found not to be acceptable, then it must be modified and reanalyzed.

16 For existing structures rehabilitation, or verification of an old infrastructure, analysis is the mostimportant component.

17 In summary, analysis is always required.

1.8 Load Transfer Elements

18 From Strength of Materials, Fig. 1.1

Figure 1.1: Types of Forces in Structural Elements (1D)

Axial: cables, truss elements, arches, membrane, shells

Flexural: Beams, frames, grids, plates

Torsional: Grids, 3D frames

Shear: Frames, grids, shear walls.

1.9 Structure Types

19 Structures can be classified as follows:



Tension & Compression Structures: only, no shear, flexure, or torsion

Cable (tension only): The high strength of steel cables, combined with the efficiency of simpletension, makes cables ideal structural elements to span large distances such as bridges, anddish roofs, Fig. 1.2

Figure 1.2: Basic Aspects of Cable Systems

Arches (mostly compression) is a “reversed cable structure”. In an arch, we seek to minimizeflexure and transfer the load through axial forces only. Arches are used for large span roofsand bridges, Fig. 1.3

Trusses have pin connected elements which can transmit axial forces only (tension and com-pression). Elements are connected by either slotted, screwed, or gusset plate connectors.However, due to construction details, there may be secondary stresses caused by relativelyrigid connections. Trusses are used for joists, roofs, bridges, electric tower, Fig. 1.4

Post and Beams: Essentially a support column on which a “beam” rests, Fig. 1.5, and 1.6.


Draft1.9 Structure Types 1–5

Figure 1.3: Basic Aspects of Arches



Figure 1.4: Types of Trusses



Figure 1.5: Variations in Post and Beams Configurations



OVERLAPPING SINGLE-STRUTCABLE-SUPPORTED BEAM

CABLE-STAYED BEAM

BRACED BEAM

VIERENDEEL TRUSS TREE-SUPPORTED TRUSS

CABLE-SUPPORTEDMULTI-STRUT

BEAM OR TRUSS

CABLE-SUPPORTED PORTAL FRAMECABLE-SUPPORTED ARCHED FRAME

SUSPENDED CABLESUPPORTED BEAM

CABLE-SUPPORTEDSTRUTED ARCH ORCABLE BEAM/TRUSS

GABLED TRUSS

BOWSTRING TRUSS

Figure 1.6: Different Beam Types



Beams: Shear, flexure and sometimes axial forces. Recall that σ = McI is applicable only for shallow

beams, i.e. span/depth at least equal to five.

Whereas r/c beams are mostly rectangular or T shaped, steel beams are usually I shaped (if thetop flanges are not properly stiffened, they may buckle, thus we must have stiffeners).

Frames: Load is co-planar with the structure. Axial, shear, flexure (with respect to one axis in 2Dstructures and with respect to two axis in 3D structures), torsion (only in 3D). The frame iscomposed of at least one horizontal member (beam) rigidly connected to vertical ones1. The verticalmembers can have different boundary conditions (which are usually governed by soil conditions).Frames are extensively used for houses and buildings, Fig. 1.7.

Figure 1.7: Basic Forms of Frames

Grids and Plates: Load is orthogonal to the plane of the structure. Flexure, shear, torsion.

In a grid, beams are at right angles resulting in a two-way dispersal of loads. Because of the rigidconnections between the beams, additional stiffness is introduced by the torsional resistance ofmembers.

Grids can also be skewed to achieve greater efficiency if the aspect ratio is not close to one.

Plates are flat, rigid, two dimensional structures which transmit vertical load to their supports.Used mostly for floor slabs.

Folded plates is a combination of transverse and longitudinal beam action. Used for long spanroofs. Note that the plate may be folded circularly rather than longitudinally. Folded plates areused mostly as long span roofs. However, they can also be used as vertical walls to support bothvertical and horizontal loads.

1The precursor of the frame structures were the Post and Lintel where the post is vertical member on which the lintelis simply posed.



Membranes: 3D structures composed of a flexible 2D surface resisting tension only. They are usuallycable-supported and are used for tents and long span roofs Fig. 1.8.

Figure 1.8: Examples of Air Supported Structures

Shells: 3D structures composed of a curved 2D surface, they are usually shaped to transmit compressiveaxial stresses only, Fig. 1.9.

Shells are classified in terms of their curvature.

1.10 Structural Engineering Courses

20 Structural engineering education can be approached from either one of two points of views:

Architectural: Start from overall design, and move toward detailed analysis.

Education: Elemental rather than global approach. Emphasis is on the individual structural elementsand not always on the total system.

CVEN3525 will seek a balance between those two approaches.

21 This is only the third of a long series of courses which can be taken in Structural Engineering, Fig.1.10


Draft1.10 Structural Engineering Courses 1–11

Figure 1.9: Basic Forms of Shells

Figure 1.10: Sequence of Structural Engineering Courses



1.11 References

22 Following are some useful references for structural engineering, those marked by † were consulted,and “borrowed from” in preparing the Lecture Notes:

Structural Art

1. Billington, D.P., The Tower and the Bridge; The new art of structural engineering, PrincetonUniversity Pres,, 1983.

Structural Engineering

1. Biggs, J.M., Introduction to Structural Engineering; Analysis and Design, Prentice Hall, 1986.

2. Gordon, J.E., Structures, or Why Things Do’nt Fall Down, Da Capo paperback, New York,1978

3. Mainstone, R., Developments in Structural Form, Allen Lane Publishers, 1975.

Structural Engineering, Architectural Analysis and Design

1. Ambrose, J., Building Structures, second Ed. Wiley, 1993.

2. Salvadori, M. and Heller, R., Structure in Architecture; The Building of Buildings, PrenticeHall, Third Edition, 1986.

3. Salvadori, M. and Levy, M., Structural Design in Architecture, Prentice hall, Second Edition,1981.

4. Salvadori, M.,Why Buildings Stand Up; The Strength of Architecture, Norton Paperack, 1990.

5. Lin, T.Y. and Stotesbury, S.D., Structural Concepts and Systems for Architects and Engineers,John Wiley, 1981.

6. † White, R. Gergely, P. and Sexmith, R., Structural Engineering; Combined Edition, JohnWiley, 1976.

7. Sandaker, B.N. and Eggen, A.P., The Structural Basis of Architecture, Whitney Library ofDesign, 1992.

Structural Analysis

1. † Arbadi, F. Structural Analysis and Behavior, McGraw-Hill, Inc., 1991.

2. Hsieh, Y.Y., Elementary Theory of Structures, Third Edition, Prentice Hall, 1988.

3. Ghali, A., and Neville, A.M., Structural Analysis, Third Edition, Chapman and Hall, 1989

Structural Design

1. † Nilson, A., and Winter, G. Design of Concrete Structures, Eleventh Edition, McGraw Hill,1991.

2. † Salmon C. and Johnson, J. Steel Structures, Third Edition, Harper Collins Publisher, 1990.

3. † Gaylord, E.H., Gaylord, C.N. and Stallmeyer, J.E., Design of Steel Structures, Third Edi-tion, McGraw Hill, 1992.

Codes

1. ACI-318-89, Building Code Requirements for Reinforced Concrete, American Concrete Insti-tute

2. Load & Resistance Factor Design, Manual of Steel Construction, American Institute of SteelConstruction.

3. Uniform Building Code, International Conference of Building Officials, 5360 South WorkmanRoad; Whittier, CA 90601

4. Minimum Design Loads in Buildings and Other Structures, ANSI A58.1, American NationalStandards Institute, Inc., New York, 1972.


Draft

Chapter 2

LOADS

2.1 Introduction

1 The main purpose of a structure is to transfer load from one point to another: bridge deck to pier;slab to beam; beam to girder; girder to column; column to foundation; foundation to soil.

2 There can also be secondary loads such as thermal (in restrained structures), differential settlementof foundations, P-Delta effects (additional moment caused by the product of the vertical force and thelateral displacement caused by lateral load in a high rise building).

3 Loads are generally subdivided into two categories

Vertical Loads or gravity load

1. dead load (DL)

2. live load (LL)

also included are snow loads.

Lateral Loads which act horizontally on the structure

1. Wind load (WL)

2. Earthquake load (EL)

this also includes hydrostatic and earth loads.

4 This distinction is helpful not only to compute a structure’s load, but also to assign different factor ofsafety to each one.

5 For a detailed coverage of loads, refer to the Universal Building Code (UBC), (UBC 1995).

2.2 Vertical Loads

6 For closely spaced identical loads (such as joist loads), it is customary to treat them as a uniformlydistributed load rather than as discrete loads, Fig. 2.1

2.2.1 Dead Load

7 Dead loads (DL) consist of the weight of the structure itself, and other permanent fixtures (such aswalls, slabs, machinery).

8 For analysis purposes, dead loads can easily be determined from the structure’s dimensions and density,Table 2.1

Draft2–2 LOADS

P P P P P P P1 2 3 4 5 6 7

TYPICAL SYSTEM OF JOISTS

SUPPORT BEAM

REPETITIVE JOIST LOADS

ACTUAL DISCRETE LOADS ON SUPPORT BEAM

ASSUMED EQUIVALENT UNIFORM LOAD

w LB/FT = TOTAL LOAD / SPAN

SPAN

Figure 2.1: Approximation of a Series of Closely Spaced Loads

Material lb/ft3 kN/m3

Aluminum 173 27.2Brick 120 18.9Concrete 145 33.8Steel 490 77.0Wood (pine) 40 6.3

Table 2.1: Unit Weight of Materials

9 For steel structures, the weight per unit length of rolled sections is given in the AISC Manual of SteelConstruction.

10 For design purposes, dead loads must be estimated and verified at the end of the design cycle. Thismakes the design process iterative.

11 Weights for building materials is given in Table 2.2

12 For preliminary design purposes the average dead loads of Table 2.3 can be used:

2.2.2 Live Loads

13 Contrarily to dead loads which are fixed and vertical, live loads (LL) are movable or moving and maybe horizontal.

14 Occupancy load may be due to people, furniture, equipment. The loads are essentially variable pointloads which can be placed anywhere.

15 In analysis load placement should be such that their effect (shear/moment) are maximized.

16 A statistical approach is used to determine a uniformly distributed static load which is equivalent tothe weight of the maximum concentration of occupants. These loads are defined in codes such as theUniform Building Code or the ANSI Code, Table 2.4.

17 For small areas (30 to 50 sq ft) the effect of concentrated load should be considered separately.

18 Since there is a small probability that the whole floor in a building be fully loaded, the UBC codespecifies that the occupancy load for members supporting an area A larger than 150 ft2 (i.e. a columnwith a total tributary area, including floors above it, larger than 150 ft2) may be reduced by R where

R = r(A − 150) ≤ 23.1(1 +

DL

LL

)(2.1)


Draft2.2 Vertical Loads 2–3

Material lb/ft2

CeilingsChannel suspended system 1Acoustical fiber tile 1

FloorsSteel deck 2-10Concrete-plain 1 in. 12Linoleum 1/4 in. 1Hardwood 4

RoofsCopper or tin 1-55 ply felt and gravel 6Shingles asphalt 3Clay tiles 9-14Sheathing wood 3Insulation 1 in. poured in place 2

PartitionsClay tile 3 in. 17Clay tile 10 in. 40Gypsum Block 5 in. 14Wood studs 2x4 (12-16 in. o.c.) 2Plaster 1 in. cement 10Plaster 1 in. gypsum 5

WallsBricks 4 in. 40Bricks 12 in. 120Hollow concrete block (heavy aggregate)4 in. 308 in. 5512 in. 80Hollow concrete block (light aggregate)4 in. 218 in. 3812 in. 55

Table 2.2: Weights of Building Materials

Material lb/ft2

Timber 40-50Steel 50-80Reinforced concrete 100-150

Table 2.3: Average Gross Dead Load in Buildings

Use or Occupancy lb/ft2

Assembly areas 50Cornices, marquees, residential balconies 60Corridors, stairs 100Garage 50Office buildings 50Residential 40Storage 125-250

Table 2.4: Minimum Uniformly Distributed Live Loads, (UBC 1995)


Draft2–4 LOADS

where r = .08 for floors, A is the supported area ( ft2) DL and LL are the dead and live loads per unitarea supported by the member. R can not exceed 40% for horizontal members and 60% for vertical ones.

Example 2-1: Live Load Reduction

In a 10 story office building with a column spacing of 16 ft in both directions, the total dead loadis 60 psf, snow load 20 psf and live load 80 psf. what is the total live load and total load for which acolumn must be designed on the ground floorSolution:

1. The tributary area is 16× 16 = 256ft2 > 150√

2. The reduction R is R = .08(16× 16− 150) = 8.48%

3. Maximum allowable reduction Rmax = 23.1(1 + 60

80

)= 40.4% which is less than 60%

√

4. The reduced cumulative load for the column of each floor is

Floor Roof 10 9 8 7 6 5 4 3 2 TotalCumulative R (%) 8.48 16.96 25.44 33.92 42.4 51.32 59.8 60 60 60Cumulative LL 20 80 80 80 80 80 80 80 80 80 740Cumulative R× LL 18.3 66.4 59.6 52.9 46.08 38.9 32.2 32 32 32 410

The resulting design live load for the bottom column has been reduced from 740 Kips to 410 Kips .

5. The total dead load is DL = (10)(60) = 600 Kips, thus the total reduction in load is 740−410740+600 ×

100= 25% .

2.2.3 Snow

19 Roof snow load vary greatly depending on geographic location and elevation. They range from20 to 45 psf, Fig. 2.2.

Figure 2.2: Snow Map of the United States, ubc

20 Snow loads are always given on the projected length or area on a slope, Fig. 2.3.



LIVE LOAD

DEAD LOAD

LENGTH RIS

E

RUN

WINDLOAD

Figure 2.3: Loads on Projected Dimensions

21 The steeper the roof, the lower the snow retention. For snow loads greater than 20 psf and roof pitchesα more than 20◦ the snow load p may be reduced by

R = (α− 20)( p40− 0.5

)(psf) (2.2)

22 Other examples of loads acting on inclined surfaces are shown in Fig. 2.4.

w1

2L

w3

w3

L1

w 2

θ

w1

w2

Snow Load

Roof Dead Load

Wind Load

1L

=Lcos θ2

Figure 2.4: Vertical and Normal Loads Acting on Inclined Surfaces


Draft2–6 LOADS

2.3 Lateral Loads

2.3.1 Wind

23 Wind load depend on: velocity of the wind, shape of the building, height, geographicallocation, texture of the building surface and stiffness of the structure.

24 Wind loads are particularly significant on tall buildings1.

25 When a steady streamline airflow of velocity V is completely stopped by a rigid body, the stagnationpressure (or velocity pressure) qs was derived by Bernouilli (1700-1782)

qs =12ρV 2 (2.3)

where the air mass density ρ is the air weight divided by the acceleration of gravity g = 32.2 ft/sec2. Atsea level and a temperature of 15oC (59oF), the air weighs 0.0765 lb/ft3 this would yield a pressure of

qs =12

(0.0765)lb/ft3

(32.2)ft/sec2

((5280)ft/mile(3600)sec/hr

V

)2(2.4)

or

qs = 0.00256V 2 (2.5)

where V is the maximum wind velocity (in miles per hour) and qs is in psf. V can be obtained fromwind maps (in the United States 70 ≤ V ≤ 110), Fig. 2.5.

Figure 2.5: Wind Map of the United States, (UBC 1995)

26 During storms, wind velocities may reach values up to or greater than 150 miles per hour, whichcorresponds to a dynamic pressure qs of about 60 psf (as high as the average vertical occupancy load inbuildings).

27 Wind pressure increases with height, Table 2.5.

28 Wind load will cause suction on the leeward sides, Fig. 2.71The primary design consideration for very high rise buildings is the excessive drift caused by lateral load (wind and

possibly earthquakes).


Draft2.3 Lateral Loads 2–7

Height Zone Wind-Velocity Map Area(in feet) 20 25 30 35 40 45 50<30 15 20 25 25 30 35 4030 to 49 20 25 30 35 40 45 5050 to 99 25 30 40 45 50 55 60100 to 499 30 40 45 55 60 70 75500 to 1199 35 45 55 60 70 80 90>1,200 40 50 60 70 80 90 100

Table 2.5: Wind Velocity Variation above Ground

Figure 2.6: Effect of Wind Load on Structures(Schueller 1996)


Draft2–8 LOADS

29 This magnitude must be modified to account for the shape and surroundings of the building. Thus,the design pressure p (psf) is given by

p = CeCqIqs (2.6)

The pressure is assumed to be normal to all walls and roofs and

Ce Velocity Pressure Coefficient accounts for height, exposure and gust factor. It accounts for thefact that wind velocity increases with height and that dynamic character of the airflow (i.e thewind pressure is not steady), Table 2.6.

Ce Exposure1.39-2.34 D Open, flat terrain facing large bodies of water1.06-2.19 C Flat open terrain, extending one-half mile or open from the site in

any full quadrant0.62-1.80 B Terrain with buildings, forest, or surface irregularities 20 ft or more

in height

Table 2.6: Ce Coefficients for Wind Load, (UBC 1995)

Cq Pressure Coefficient is a shape factor which is given in Table 2.7 for gabled frames.

Windward Side Leeward SideGabled Frames (V:H)

Roof Slope <9:12 −0.7 −0.79:12 to 12:12 0.4 −0.7>12:12 0.7 −0.7

Walls 0.8 −0.5Buildings (height < 200 ft)

Vertical Projections height < 40 ft 1.3 −1.3height > 40 ft 1.4 −1.4

Horizontal Projections −0.7 −0.7

Table 2.7: Wind Pressure Coefficients Cq, (UBC 1995)

I Importance Factor as given by Table 2.8. where

Importance Factor IOccupancy Category Earthquake Wind

I Essential facilities 1.25 1.15II Hazardous facilities 1.25 1.15III Special occupancy structures 1.00 1.00IV Standard occupancy structures 1.00 1.00

Table 2.8: Importance Factors for Wind and Earthquake Load, (UBC 1995)

I Essential Facilities: Hospitals; Fire and police stations; Tanks; Emergency vehicle shelters,standby power-generating equipment; Structures and equipment in government. communica-tion centers.

II Hazardous Facilities: Structures housing, supporting or containing sufficient quantities oftoxic or explosive substances to be dangerous to the safety of the general public if released.



III Special occupancy structure: Covered structures whose primary occupancy is public as-sembly, capacity > 300 persons.Buildings for schools through secondary or day-care centers, capacity > 250 persons.Buildings for colleges or adult education schools, capacity > 500 persons.Medical facilities with 50 or more resident incapacitated patients, but not included aboveJails and detention facilitiesAll structures with occupancy >5,000 persons.Structures and equipment in power generating stations and other public utility facilities notincluded above, and required for continued operation.

IV Standard occupancy structure: All structures having occupancies or functions not listedabove.

30 For the preliminary design of ordinary buildings Ce = 1.0 and Cq = 1.3 may be assumed, yielding

p = (1.3).020256V 2 = .00333V 2 (2.7)

which corresponds to a pressure of 21 psf for a wind speed of 80 mph, Fig. 2.7, Table 2.9.

ExposureHeight B CAbove Basic Wind Speed (mph)Grade (ft) 70 80 70 800-15 10 13 17 2320 11 14 18 2425 12 15 19 2530 12 16 20 2640 14 18 21 2860 17 22 25 3380 18 24 27 35100 20 26 28 37120 21 28 29 38160 23 30 31 41200 25 33 33 43300 29 37 36 47400 32 41 38 50

Table 2.9: Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Structures

Example 2-2: Wind Load

Determine the wind forces on the building shown on below which is built in St Louis and is surroundedby trees.Solution:

1. From Fig. 2.5 the maximum wind velocity is St. Louis is 70 mph, since the building is protectedwe can take Ce = 0.7, I = 1.. The base wind pressure is qs = 0.00256× (70)2 = 12.54 psf.


Draft2–10 LOADS

0 5 10 15 20 25 30 35 40 45 50Approximate Design Wind Pressure (psf)

0

50

100

150

200

250

300

350

400

Hei

ght A

bove

Gra

de (

ft)

Exposure B, 70 mphExposure B, 80 mphExposure C, 70 mphExposure C, 80 mph

Figure 2.7: Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Structures

2. The slope of the roof is 8:15=6.4:12 which gives Cq = −0.7 for both the windward and leewardsides. The vertical walls have Cq = 0.8 for the windward side and Cq = −0.5 for the leeward one.

3. Thus the applied pressure on the roof is p = 0.7× (−0.7)× 12.54 = -6.14 psf that is the roof issubjected to uplift.

4. The winward wall, the pressure is 0.7 × 0.8 × 12.54 = 7.02 psf , and for the leeward wall 0.7 ×(−0.5)× 12.54 = -4.39 psf (suction) ,

5. The direction of the wind can change and hence each structural component must be designed toresist all possible load combinations.

6. For large structures which may be subjected to large wind loads, testing in a wind tunnel of thestructure itself and its surroundings is often accomplished.



2.3.2 Earthquakes

31 Buildings should be able to resist

Minor earthquakes without damage

Moderate earthquakes without structural damage but possibly with some nonstructural damages

Major earthquakes without collapse but possibly with some structural damage as well as nonstruc-tural damage.

This is achieved through an appropriate dynamic analysis.

32 For preliminary designs or for small structures an equivalent horizontal static load can be deter-mined.

33 Actual loads depend on the following

1. Intensity of the ground acceleration (including soil/rock properties).

2. Dynamic properties of the building, such as its mode shapes and periods of vibration and itsdamping characteristics.

3. Mass of the building.

34 A critical factor in the dynamic response of a structure is the fundamental period of the structure’svibration (or first mode of vibration). This is the time required for one full cycle of motion, Fig. 2.8. Ifthe earthquake excitation has a frequency close to the one of the building, then resonance may occur.This should be avoided.

Figure 2.8: Vibrations of a Building

35 Earthquake load manifests itself as a horizontal force due to the (primarily) horizontal inertia force(F = ma).

36 The horizontal force at each level is calculated as a portion of the base shear force V

V =ZIC

RWW (2.8)


Draft2–12 LOADS

where:

Z: Zone Factor: to be determined from Fig. 2.9 and Table 2.10.

Figure 2.9: Seismic Zones of the United States, (UBC 1995)

Seismic Zone 0 1 2A 2B 3 4Z 0 0.075 0.15 0.2 0.3 0.4

Table 2.10: Z Factors for Different Seismic Zones, ubc

I: Importance Factor: which was given by Table 2.8.

C: Design Response Spectrum given by

C =1.25ST 2/3

≤ 2.75 (2.9)

T is the fundamental period of vibration of the building in seconds. This can be determined fromeither the free vibration analysis of the building, or estimated from the following empirical formula

T = Ct(hn)3/4 (2.10)

where:hn is the building height above base in ft.

andCt 0.035 steel moment resisting framesCt 0.030 reinforced concrete moment resisting frames and eccentrically braced framesCt 0.020 all other buildings

S: Site

Coefficient given by Table 2.11 Note that most of the damages in the 1990? earthquake in SanFrancisco occurred in the marina where many houses were built on soft soil.

and

C

RW≥ 0.075 (2.11)



Type Description S FactorS1 A soil profile with either rock-like material or stiff/dense soil less

than 200 ft.1.0

S2 Dense or stiff soil exceeding 200 ft 1.2S3 70 ft or more soil containing more than 20 ft of soft to medium stiff

clay but not more than 40 ft. of soft clay.1.5

S4 Soil containing more than 40 ft of soft clay 2.0

Table 2.11: S Site Coefficients for Earthquake Loading, (UBC 1995)

Structural System RW H (ft)Bearing wall system

Light-framed walls with shear panelsPlywood walls for structures three stories or less 8 65All other light-framed walls 6 65

Shear wallsConcrete 8 240Masonry 8 160

Building frame system using trussing or shear walls)

Steel eccentrically braced ductiel frame 10 240Light-framed walls with shear panels

Plywood walls for structures three stories or less 9 65All other light-framed walls 7 65

Shear wallsConcrete 8 240Masonry 8 160

Concentrically braced framesSteel 8 160Concrete (only for zones I and 2) 8 -Heavy timber 8 65

Moment-resisting frame systemSpecial moment-resisting frames (SMRF)

Steel 12 N.L.Concrete 12 N.L.

Concrete intermediate moment-resisting frames (IMRF)only for zones 1 and 2 8 -Ordinary moment-resisting frames (OMRF)

Steel 6 160Concrete (only for zone 1) 5 -

Dual systems (selected cases are for ductile rigid frames only)Shear wallsConcrete with SMRF 12 N.L.Masonry with SMRF 8 160

Steel eccentrically braced ductile frame 6-12 160-N.L.Concentrically braced frame 12 N. L.

Steel with steel SMRF 10 N.L.Steel with steel OMRF 6 160Concrete with concrete SMRF (only for zones 1 and 2) 9 -

Table 2.12: Partial List of RW for Various Structure Systems, (UBC 1995)


Draft2–14 LOADS

RW is given by Table 2.12.

W Load total structure load.

37 The horizontal force V is distributed over the height of the building in two parts. The first (appliedonly if T ≥ 0.7 sec.) is a concentrated force F1 equal to

Ft = 0.07TV ≤ 0.25V (2.12)

is applied at the top of the building due to whiplash. The balance of the force V − Ft is distributed asa triangular load diminishing to zero at the base.

38 Assuming a floor weight constant for every floor level, then the force acting on each one is given by

Fx =(V − Ft)hx

h1 + h2 + · · ·+ hn =(V − Ft)hx

Σni=1hi(2.13)

where hi and hn are the height in ft above the base to level i, n or x respectively

Example 2-3: Earthquake Load on a Frame

Determine the approximate earthquake forces for the ductile hospital frame structure shown below.The DL for each floor is 200 lb/ft and the LL is 400 lb/ft. The structure is built on soft soil. Use DLplus 50%LL as the weight of each floor. The building is in zone 3.

Solution:

1. The fundamental period of vibration is

T = Ct(hn)3/4 = (0.030)(24)3/4 = 0.32 sec. (2.14)

2. The C coefficient is

C =1.25ST 2/3

=(1.25)(2.0)(0.32)2/3

= 5.344 > 2.75 (2.15)

use C = 2.75.

3. The other coefficients are: Z =0.3; I=1.25; RW=12

4. CheckC

RW=

2.7512

= 0.23 > 0.075√

(2.16)

5. The total vertical load is

W = 2 ((200 + 0.5(400)) (20) = 16000 lbs (2.17)



6. The total seismic base shear is

V =ZIC

RW=

(0.3)(1.25)(2.75)12

= 0.086W (2.18-a)

= (0.086)(16000) = 1375 lbs (2.18-b)

7. Since T < 0.7 sec. there is no whiplash.

8. The load on each floor is thus given by

F2 =(1375)(24)12 + 24

= 916.7 lbs (2.19-a)

F1 =(1375)(12)12 + 24

= 458.3 lbs (2.19-b)

Example 2-4: Earthquake Load on a Tall Building, (Schueller 1996)

Determine the approximate critical lateral loading for a 25 storey, ductile, rigid space frame concretestructure in the short direction. The rigid frames are spaced 25 ft apart in the cross section and 20ft in the longitudinal direction. The plan dimension of the building is 175x100 ft, and the structure is25(12)=300 ft high. This office building is located in an urban environment with a wind velocity of 70mph and in seismic zone 4. For this investigation, an average building total dead load of 192 psf is used.Soil conditions are unknown.

470 k

2638 k

1523 k

2(30

0)/3

=20

0’

300/

2=15

0’

25(1

2)=

300’

84000 k

3108 k

5(20)=100’

7(25

)=17

5’

Solution:

1. The total building weight is

W = (0.1926) psf(100× 175) ft2 × 25 storeys = 84, 000 k (2.20)


Draft2–16 LOADS

2. the fundamental period of vibration for a rigid frame is

T = Ct(hn)3/4 = 0.030(300)3/4 = 2.16 sec. > 0.7 sec.√

(2.21)

3. The C coefficient is

C =1.25ST 2/3

=(1.25)(1.5)(2.16)2/3

= 1.12 ≤ 2.75√

(2.22)

4. The other coefficients are Z=0.4; I=1, RW=12

5. We checkC

RW=

1.1212

= 0.093 ≥ 0.075√

(2.23)

6. The total seismic base shear along the critical short direction is

V =ZIC

RWW =

(0.4)(1)(1.12)(12)

W = 0.037W (2.24-a)

= (0.037)(84000) = 3108 kip (2.24-b)

7. Since T > 0.7 sec., the whiplash effect must be considered

Ft = 0.07TV = (0.07)(2.16)(3108) = 470 k ✛ (2.25-a)le 0.25V = (0.25)(3108) = 777 k (2.25-b)

Hence the total triangular load is

V − Ft = 3108− 470 = 2638 k (2.26)

8. let us check if wind load governs. From Table xx we conservatively assume a uniform wind pressureof 29 psf resulting in a total lateral force of

PW = (0.029) psf(175× 300) ft2 = 1523 k < 3108 k (2.27)

The magnitude of the total seismic load is clearly larger than the total wind force.

2.4 Other Loads

2.4.1 Hydrostatic and Earth

39 Structures below ground must resist lateral earth pressure.

q = Kγh (2.28)

where γ is the soil density, K = 1−sinΦ1+sinΦ is the pressure coefficient, h is the height.

40 For sand and gravel γ = 120 lb/ ft3, and Φ ≈ 30◦.

41 If the structure is partially submerged, it must also resist hydrostatic pressure of water, Fig. 2.10.

q = γWh (2.29)

where γW = 62.4 lbs/ft3.

Example 2-5: Hydrostatic Load


Draft2.4 Other Loads 2–17

Figure 2.10: Earth and Hydrostatic Loads on Structures

The basement of a building is 12 ft below grade. Ground water is located 9 ft below grade, whatthickness concrete slab is required to exactly balance the hydrostatic uplift?Solution:The hydrostatic pressure must be countered by the pressure caused by the weight of concrete. Since p =γh we equate the two pressures and solve for h the height of the concrete slab (62.4) lbs/ft3 × (12− 9) ft︸︷︷︸

water

=

(150) lbs/ft3 × h︸︷︷︸concrete

⇒ h = (62.4) lbs/ft3

(150) lbs/ft3(3) ft(12) in/ft = 14.976 in 15.0 inch

2.4.2 Thermal

42 If a member is uniformly heated (or cooled) without restraint, then it will expand (or contract).This expansion is given by

∆l = αl∆T (2.30)

where α is the coefficient of thermal expansion, Table 2.13

α (/◦F )Steel 6.5× 10−6

Concrete 5.5× 10−6

Table 2.13: Coefficients of Thermal Expansion

43 If the member is restrained against expansion, then a compressive stress σ = Eα∆T is developed.

44 To avoid excessive stresses due to thermal loading expansion joints are used in bridges and buildings.

Example 2-6: Thermal Expansion/Stress (Schueller 1996)

A low-rise building is enclosed along one side by a 100 ft-long clay masonry (α = 3.6×10−6 in./in./oF,E = 2, 400, 000 psi) bearing wall. The structure was built at a temperature of 60oF and is located inthe northern part of the United States where the temperature range is between -20o and +120oF.Solution:


Draft2–18 LOADS

1. Assuming that the wall can move freely with no restraint from cross-walls and foundation, the wallexpansion and contraction (summer and winter) are given by

∆LSummer = α∆TL = (3.6× 10−6) in/ in/oF (120− 60)oF (100) ft(12) in/ft = 0.26 in(2.31-a)

∆LWinter = α∆TL = (3.6× 10−6) in/ in/oF (−20− 60)oF (100) ft(12) in/ft = -0.35 in(2.31-b)

2. We now assume (conservatively) that the free movement cannot occur (∆L = 0) hence the resultingstress would be equal to σ = Eε = E∆L

L = E α∆TLL = Eα∆T

σSummer = Eα∆T = (2, 400, 000)lbs

in2(3.6× 10−6) in/ in/oF (120− 60)oF = 518

lbs

in2Tension(2.32-a)

σWinter = Eα∆T = (2, 400, 000)lbs

in2(3.6× 10−6) in/ in/oF (−20− 60)oF = -691

lbs

in2Compression(2.32-b)

(2.32-c)

Note that the tensile stresses being beyond the masonry capacity, cracking will occur.

2.4.3 Bridge Loads

For highway bridges, design loads are given by the AASHTO (Association of American State HighwayTransportation Officials). The HS-20 truck is used for the design of bridges on main highways, Fig. 6.5.Either the design truck with specified axle loads and spacing must be used or the equivalent uniformload and concentrated load. This loading must be placed such that maximum stresses are produced.

Figure 2.11: Truck Load

2.4.4 Impact Load

2.5 Other Important Considerations

2.5.1 Load Combinations

45 Live loads specified by codes represent the maximum possible loads.


Draft2.5 Other Important Considerations 2–19

46 The likelihood of all these loads occurring simultaneously is remote. Hence, building codes allowcertain reduction when certain loads are combined together.

47 Furthermore, structures should be designed to resist a combination of loads.

48 Denoting D= dead; L= live; Lr= roof live; W= wind; E= earthquake; S= snow; T= temperature;H= soil:

49 For the load and resistance factor design (LRFD) method of concrete structures, the American Con-crete Institute (ACI) Building design code (318) (318 n.d.) requires that the following load combinationsbe considered:

1. 1.4D+1.7L

2. 0.75(1.4D+1.7L+1.7W)

3. 0.9D+1.3W

4. 1.4D +1.7L+1.7H

5. 0.75(1.4D+1.4T+1.7L)

6. 1.4(D+T)

whereas for steel structures, the American Institute of Steel Construction (AISC) code, (of Steel COn-struction 1986) requires that the following combinations be verified

1. 1.4D

2. 1.2D+1.6L+0.5(Lr or S)

3. 1.2D+0.5L (or 0.8W)+1.6(Lr or S)

4. 1.2D+0.5L+0.5(Lr or S)+1.3W

5. 1.2D+0.5L(or 0.2 S)+1.5E

6. 0.9D+1.3W(or 1.5 E)

50 Analysis can be separately performed for each of the basic loads (L, D, W, etc) and then using theprinciple of superposition the loads can be linearly combined (unless the elastic limit has been reached).

51 Loads are often characterized as Usual, Unusual and Extreme.

2.5.2 Load Placement

52 Only the dead load is static. The live load on the other hand may or may not be applied on a givencomponent of a structure. Hence, the load placement arrangement resulting in the highest internal forces(moment +ve or -ve, shear) at different locations must be considered, Fig. 2.12.

2.5.3 Structural Response

53 Under the action of the various forces and loadings described above, the structure must be able torespond with proper behavior, Fig. 7.1.


Draft2–20 LOADS

Figure 2.12: Load Placement to Maximize Moments

Figure 2.13: Load Life of a Structure, (Lin and Stotesbury 1981)



��

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��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

Figure 2.14: Concept of Tributary Areas for Structural Member Loading

a

b b

one way slab

a/b > 2 a/b <2

Two way slab

Figure 2.15: One or Two Way actions in Slabs


Draft2–22 LOADS

2.5.4 Tributary Areas

54 For preliminary analysis, the tributary area of a structural component will determine the total appliedload.

55 For a slab simply supported over four linear supports, we may have a one way or tow way action ifthe ratio a/b is greater or smaller than two respectively, Fig. 2.15.

56 Whereas we will be focusing on the design of a reinforced concrete or steel section, we must keep inmind the following:

1. The section is part of a beam or girder.

2. The beam or girder is really part of a three dimensional structure in which load is transmittedfrom any point in the structure to the foundation through any one of various structural forms.

57 Load transfer in a structure is accomplished through a “hierarchy” of simple flexural elements whichare then connected to the columns, Fig. 2.16 or by two way slabs as illustrated in Fig. 2.17.

58 An example of load transfer mechanism is shown in Fig. 2.18.



Figure 2.16: Load Transfer in R/C Buildings


Draft2–24 LOADS

Figure 2.17: Two Way Actions

��

��

��

��

A B C D

HG F E

a a a

b

LL=40lbs/ft 2

b=12’; a=4’; b/a=12/4=3>2 One way slab

(40) lb/ft (4) ft= 160 lb/ft=0.16 k/ft2

(0.16)(12)/2=0.96 k0.96 k

12 ft

4 ft 4 ft 4 ft

0.48 k0.96 k 0.96 k

0.48 k

A B C D

0.48+0.96=1.44 k

1.44 k

b=10’; a=6’; b/a<2 Two way slab

3’

4’

3’

3’ 3’

G

A B

H

3’ 4’ 3’(0.24)(2+3/2)=0.84k

(2)(40)lb/ft (3)ft=0.24 k/ft2

0.84 k

B G

0.42k 0.42k0.84k0.84k (40)lb/ft (3)ft=0.12k/ft

2

3’ 3’ 3’ 3’ 3’ 3’

(0.42)+(0.84)+(0.12)[6/2+3/2]=1.8k

1.8k1.8k

AB C

D

B G

Figure 2.18: Example of Load Transfer


Draft

Chapter 3

STRUCTURAL MATERIALS

1 Proper understanding of structural materials is essential to both structural analysis and to structuraldesign.

2 Characteristics of the most commonly used structural materials will be highlighted.

3.1 Steel

3.1.1 Structural Steel

3 Steel is an alloy of iron and carbon. Its properties can be greatly varied by altering the carbon content(always less than 0.5%) or by adding other elements such as silicon, nickel, manganese and copper.

4 Practically all grades of steel have a Young Modulus equal to 29,000 ksi, density of 490 lb/cu ft, anda coefficient of thermal expansion equal to 0.65× 10−5 /deg F.

5 The yield stress of steel can vary from 40 ksi to 250 ksi. Most commonly used structural steel are A36(σyld = 36 ksi) and A572 (σyld = 50 ksi), Fig. 3.6

Figure 3.1: Stress Strain Curves of Concrete and Steel

6 Structural steel can be rolled into a wide variety of shapes and sizes. Usually the most desirablemembers are those which have a large section moduli (S) in proportion to their area (A), Fig. 3.2.

Draft3–2 STRUCTURAL MATERIALS

Figure 3.2: Standard Rolled Sections

7 Steel can be bolted, riveted or welded.

8 Sections are designated by the shape of their cross section, their depth and their weight. For exampleW 27× 114 is a W section, 27 in. deep weighing 114 lb/ft.

9 Common sections are:

S sections were the first ones rolled in America and have a slope on their inside flange surfaces of 1 to6.

W or wide flange sections have a much smaller inner slope which facilitates connections and rivetting.W sections constitute about 50% of the tonnage of rolled structural steel.

C are channel sections

MC Miscellaneous channel which can not be classified as a C shape by dimensions.

HP is a bearing pile section.

M is a miscellaneous section.

L are angle sections which may have equal or unequal sides.

WT is a T section cut from a W section in two.

Properties of structural steel are tabulated in Table 3.1.

10 Rolled sections, Fig. 3.3 and welded ones, Fig 3.4 have residual stresses. Those originate duringthe rolling or fabrication of a member. The member is hot just after rolling or welding, it cools unevenlybecause of varying exposure. The area that cool first become stiffer, resist contraction, and developcompressive stresses. The remaining regions continue to cool and contract in the plastic condition anddevelop tensile stresses.

11 Due to those residual stresses, the stress-strain curve of a rolled section exhibits a non-linear segmentprior to the theoretical yielding, Fig. 3.5. This would have important implications on the flexural andaxial strength of beams and columns.

3.1.2 Reinforcing Steel

12 Steel is also used as reinforcing bars in concrete, Table 3.2. Those bars have a deformation on theirsurface to increase the bond with concrete, and usually have a yield stress of 60 ksi1.

1Stirrups which are used as vertical reinforcement to resist shear usually have a yield stress of only 40 ksi.


Draft3.1 Steel 3–3

ASTMDesig.

Shapes Available Use σy (ksi) σu (ksi)

A36 Shapes and bars Riveted, bolted, welded;Buildings and bridges

36 up through 8 in. (32 above8.)

A500 Cold formed welded andseamless sections;

General structural pur-pose Riveted, welded orbolted;

Grade A: 33; Grade B: 42;Grade C: 46

A501 Hot formed welded and seam-less sections;

Bolted and welded 36

A529 Plates and bars 12in and less

thick;Building frames andtrusses; Bolted andwelded

42

A606 Hot and cold rolled sheets; Atmospheric corrosionresistant

45-50

A611 Cold rolled sheet in cutlengths

Cold formed sections Grade C 33; Grade D 40;Grade E 80

A 709 Structural shapes, plates andbars

Bridges Grade 36: 36 (to 4 in.); Grade50: 50; Grade 100: 100 (to2.5in.) and 90 (over 2.5 to 4in.)

Table 3.1: Properties of Major Structural Steels

Compression (-)

Tension (+)(+)

(-)

Maximum compressive

stress, say 12 ksi average

Figure 3.3: Residual Stresses in Rolled Sections

-

+ +

+

_

+

+

-

-

say 12 ksi

20 ksi

Welded H

say 35 ksitension

say 20 ksi

say 40 ksi

Welded box

say 20 ksicompression

Figure 3.4: Residual Stresses in Welded Sections



.

..

Maximumresidualcompressivestress

no residual stressIdeal coupon containing

Members withresidual stress

F

F

y

p

2

1

3

1

2 3Average copressive strain

Ave

rage

str

ess

P/A

Shaded portion indicates areawhich has achieved a stress Fy

Figure 3.5: Influence of Residual Stress on Average Stress-Strain Curve of a Rolled Section

Bar Designation Diameter Area Perimeter Weight(in.) ( in2) in lb/ft

No. 2 2/8=0.250 0.05 0.79 0.167No. 3 3/8=0.375 0.11 1.18 0.376No. 4 4/8=0.500 0.20 1.57 0.668No. 5 5/8=0.625 0.31 1.96 1.043No. 6 6/8=0.750 0.44 2.36 1.5202No. 7 7/8=0.875 0.60 2.75 2.044No. 8 8/8=1.000 0.79 3.14 2.670No. 9 9/8=1.128 1.00 3.54 3.400No. 10 10/8=1.270 1.27 3.99 4.303No. 11 11/8=1.410 1.56 4.43 5.313No. 14 14/8=1.693 2.25 5.32 7.650No. 18 18/8=2.257 4.00 7.09 13.60

Table 3.2: Properties of Reinforcing Bars


Draft3.2 Aluminum 3–5

13 Steel loses its strength rapidly above 700 deg. F (and thus must be properly protected from fire), andbecomes brittle at −30 deg. F2.

14 Steel is also used as wire strands and ropes for suspended roofs, cable-stayed bridges, fabric roofs andother structural applications. A strand is a helical arrangement of wires around a central wire. A ropeconsists of multiple strands helically wound around a central plastic core, and a modulus of elasticity of20,000 ksi, and an ultimate strength of 220 ksi.

15 Prestressing Steel cables have an ultimate strength up to 270 ksi.

3.2 Aluminum

16 Aluminum is used whenever light weight combined with strength is an important factor. Thoseproperties, along with its resistance to corrosion have made it the material of choice for airplanestructures, light roof framing.

17 Aluminum members can be connected by riveting, bolting and to a lesser extent by welding.

18 Aluminum has a modulus of elasticity equal to 10,000 ksi (about three times lower than steel),a coefficient of thermal expansion of 2.4× 10−5 and a density of 173 lbs/ft3.

19 The ultimate strength of pure aluminum is low (13,000 psi) but with the addition of alloys it can goup.

20 When aluminum is in contact with other metals in the presence of an electrolyte, galvanic corrosionmay cause damage. Thus, steel and aluminum in a structure must be carefully separated by means ofpainting or a nonconductive material.

3.3 Concrete

21 Concrete is a mixture of Portland cement3, water, and aggregates (usually sand and crushed stone).An ideal mixture is one in which:

1. A minimum amount of cement-water paste is used to fill the interstices between the particles ofaggregates.

2. A minimum amount of water is provided to complete the chemical reaction with cement.

In such a mixture, about 3/4 of the volume is constituted by the aggregates, and the remaining 1/4being the cement paste.

22 Smaller particles up to 1/4 in. in size are called fine aggregates, and the larger ones being coarseaggregates.

23 Contrarily to steel to modulus of elasticity of concrete depends on the strength and is given by

E = 57, 000√f ′c (3.1)

orE = 33γ1.5

√f ′c (3.2)

where both f ′c and E are in psi and γ is in lbs/ft3.

24 Typical concrete (compressive) strengths range from 3,000 to 6,000 psi; However high strengthconcrete can go up to 14,000 psi.

25 All concrete fail at an ultimate strain of 0.003, Fig. 3.6.

26 Pre-peak nonlinearity is caused by micro-cracking Fig. 3.7.2This brittleness is even more accentuated in high strength steel, and has been the cause of some bridge collapses during

winter nights. (CVEN 6831 Fracture Mechanics)3Portland cement is a mixture of calcareous and argillaceous materials which are calcined in a kiln and then pulverized.

When mixed with water, cement hardens through a process called hydration.



Figure 3.6: Concrete Stress-Strain curve

εu

f’c

.5f’c

linear

non-linear

Figure 3.7: Concrete microcracking


Draft3.4 Masonry 3–7

27 The tensile strength of concrete f ′t is about 10% of the compressive strength.

28 Density of normal weight concrete is 145 lbs/ft3 and 100 lbs/ft3 for lightweight concrete.

29 Coefficient of thermal expansion is 0.65× 10−5 /deg F for normal weight concrete.

30 When concrete is poured (or rather placed), the free water not needed for the hydration processevaporates over a period of time and the concrete will shrink. This shrinkage is about 0.05% after oneyear (strain). Thus if the concrete is restrained, then cracking will occur4.

31 Concrete will also deform with time due to the applied load, this is called creep. This should be takeninto consideration when computing the deflections (which can be up to three times the instantaneouselastic deflection).

3.4 Masonry

32 Masonry consists of either natural materials, such as stones, or of manufactured products such asbricks and concrete blocks5, stacked and bonded together with mortar.

33 As for concrete, all modern structural masonry blocks are essentially compression members with lowtensile resistance.

34 The mortar used is a mixture of sand, masonry cement, and either Portland cement or hydrated lime.

3.5 Timber

35 Timber is one of the earliest construction materials, and one of the few natural materials with goodtensile properties.

36 The properties of timber vary greatly, and the strength is time dependent.

37 Timber is a good shock absorber (many wood structures in Japan have resisted repeated earthquakes).

38 The most commonly used species of timber in construction are Douglas fir, southern pine, hemlockand larch.

39 Members can be laminated together under good quality control, and flexural strengths as high as2,500 psi can be achieved.

3.6 Steel Section Properties

40 Dimensions and properties of rolled sections are tabulated in the following pages, Fig. 3.8.

4For this reason a minimum amount of reinforcement is always necessary in concrete, and a 2% reinforcement, canreduce the shrinkage by 75%.

5Mud bricks were used by the Babylonians, stones by the Egyptians, and ice blocks by the Eskimos...



Figure 3.8: W and C sections


Draft3.6 Steel Section Properties 3–9

Designation A d tw bf tf

bf2tf

hctw

Ix Sx rx Iy Sy ry Zx Zy J

in2 in in in in in4 in3 in in4 in3 in in3 in3 in4

W 44x285 83.8 44.02 1.025 11.810 1.770 3.3 38.0 24600 1120 17.1 490 83 2.42 1310.0 135.0 60.0W 44x248 72.8 43.62 0.865 11.810 1.575 3.7 44.9 21400 983 17.2 435 74 2.44 1150.0 118.0 40.7W 44x224 65.8 43.31 0.785 11.810 1.415 4.2 49.4 19200 889 17.1 391 66 2.44 1030.0 105.0 30.0W 44x198 58.0 42.91 0.710 11.810 1.220 4.8 54.9 16700 776 16.9 336 57 2.41 902.0 90.0 20.1W 40x328 96.4 40.00 0.910 17.910 1.730 5.2 37.6 26800 1340 16.7 1660 185 4.15 1510.0 286.0 74.2W 40x298 87.6 39.69 0.830 17.830 1.575 5.7 41.2 24200 1220 16.6 1490 167 4.12 1370.0 257.0 56.3W 40x268 78.8 39.37 0.750 17.750 1.415 6.3 45.6 21500 1090 16.5 1320 149 4.09 1220.0 229.0 41.1W 40x244 71.7 39.06 0.710 17.710 1.260 7.0 48.1 19200 983 16.4 1170 132 4.04 1100.0 203.0 30.4W 40x221 64.8 38.67 0.710 17.710 1.065 8.3 48.1 16600 858 16.0 988 112 3.90 967.0 172.0 21.2W 40x192 56.5 38.20 0.710 17.710 0.830 10.7 48.1 13500 708 15.5 770 87 3.69 807.0 135.0 13.7W 40x655 192.0 43.62 1.970 16.870 3.540 2.4 17.4 56500 2590 17.2 2860 339 3.86 3060.0 541.0 596.0W 40x593 174.0 42.99 1.790 16.690 3.230 2.6 19.1 50400 2340 17.0 2520 302 3.81 2750.0 481.0 451.0W 40x531 156.0 42.34 1.610 16.510 2.910 2.8 21.2 44300 2090 16.9 2200 266 3.75 2450.0 422.0 329.0W 40x480 140.0 41.81 1.460 16.360 2.640 3.1 23.4 39500 1890 16.8 1940 237 3.72 2180.0 374.0 245.0W 40x436 128.0 41.34 1.340 16.240 2.400 3.4 25.5 35400 1710 16.6 1720 212 3.67 1980.0 334.0 186.0W 40x397 116.0 40.95 1.220 16.120 2.200 3.7 28.0 32000 1560 16.6 1540 191 3.65 1790.0 300.0 142.0W 40x362 106.0 40.55 1.120 16.020 2.010 4.0 30.5 28900 1420 16.5 1380 173 3.61 1630.0 270.0 109.0W 40x324 95.3 40.16 1.000 15.905 1.810 4.4 34.2 25600 1280 16.4 1220 153 3.57 1460.0 239.0 79.4W 40x297 87.4 39.84 0.930 15.825 1.650 4.8 36.8 23200 1170 16.3 1090 138 3.54 1330.0 215.0 61.2W 40x277 81.3 39.69 0.830 15.830 1.575 5.0 41.2 21900 1100 16.4 1040 132 3.58 1250.0 204.0 51.1W 40x249 73.3 39.38 0.750 15.750 1.420 5.5 45.6 19500 992 16.3 926 118 3.56 1120.0 182.0 37.7W 40x215 63.3 38.98 0.650 15.750 1.220 6.5 52.6 16700 858 16.2 796 101 3.54 963.0 156.0 24.4W 40x199 58.4 38.67 0.650 15.750 1.065 7.4 52.6 14900 769 16.0 695 88 3.45 868.0 137.0 18.1W 40x183 53.7 38.98 0.650 11.810 1.220 4.8 52.6 13300 682 15.7 336 57 2.50 781.0 89.6 19.6W 40x167 49.1 38.59 0.650 11.810 1.025 5.8 52.6 11600 599 15.3 283 48 2.40 692.0 76.0 14.0W 40x149 43.8 38.20 0.630 11.810 0.830 7.1 54.3 9780 512 14.9 229 39 2.29 597.0 62.2 9.6W 36x848 249.0 42.45 2.520 18.130 4.530 2.0 12.5 67400 3170 16.4 4550 501 4.27 3830.0 799.0 1269.0W 36x798 234.0 41.97 2.380 17.990 4.290 2.1 13.2 62600 2980 16.4 4200 467 4.24 3570.0 743.0 1073.0W 36x720 211.0 41.19 2.165 17.775 3.900 2.3 14.5 55300 2690 16.2 3680 414 4.18 3190.0 656.0 804.0W 36x650 190.0 40.47 1.970 17.575 3.540 2.5 16.0 48900 2420 16.0 3230 367 4.12 2840.0 580.0 600.0W 36x588 172.0 39.84 1.790 17.400 3.230 2.7 17.6 43500 2180 15.9 2850 328 4.07 2550.0 517.0 453.0W 36x527 154.0 39.21 1.610 17.220 2.910 3.0 19.6 38300 1950 15.8 2490 289 4.02 2270.0 454.0 330.0W 36x485 142.0 38.74 1.500 17.105 2.680 3.2 21.0 34700 1790 15.6 2250 263 3.98 2070.0 412.0 260.0W 36x439 128.0 38.26 1.360 16.965 2.440 3.5 23.1 31000 1620 15.6 1990 235 3.95 1860.0 367.0 195.0W 36x393 115.0 37.80 1.220 16.830 2.200 3.8 25.8 27500 1450 15.5 1750 208 3.90 1660.0 325.0 143.0W 36x359 105.0 37.40 1.120 16.730 2.010 4.2 28.1 24800 1320 15.4 1570 188 3.87 1510.0 292.0 109.0W 36x328 96.4 37.09 1.020 16.630 1.850 4.5 30.9 22500 1210 15.3 1420 171 3.84 1380.0 265.0 84.5W 36x300 88.3 36.74 0.945 16.655 1.680 5.0 33.3 20300 1110 15.2 1300 156 3.83 1260.0 241.0 64.2W 36x280 82.4 36.52 0.885 16.595 1.570 5.3 35.6 18900 1030 15.1 1200 144 3.81 1170.0 223.0 52.6W 36x260 76.5 36.26 0.840 16.550 1.440 5.7 37.5 17300 953 15.0 1090 132 3.78 1080.0 204.0 41.5W 36x245 72.1 36.08 0.800 16.510 1.350 6.1 39.4 16100 895 15.0 1010 123 3.75 1010.0 190.0 34.6W 36x230 67.6 35.90 0.760 16.470 1.260 6.5 41.4 15000 837 14.9 940 114 3.73 943.0 176.0 28.6W 36x256 75.4 37.43 0.960 12.215 1.730 3.5 33.8 16800 895 14.9 528 86 2.65 1040.0 137.0 53.3W 36x232 68.1 37.12 0.870 12.120 1.570 3.9 37.3 15000 809 14.8 468 77 2.62 936.0 122.0 39.8W 36x210 61.8 36.69 0.830 12.180 1.360 4.5 39.1 13200 719 14.6 411 68 2.58 833.0 107.0 28.0W 36x194 57.0 36.49 0.765 12.115 1.260 4.8 42.4 12100 664 14.6 375 62 2.56 767.0 97.7 22.2W 36x182 53.6 36.33 0.725 12.075 1.180 5.1 44.8 11300 623 14.5 347 58 2.55 718.0 90.7 18.4W 36x170 50.0 36.17 0.680 12.030 1.100 5.5 47.8 10500 580 14.5 320 53 2.53 668.0 83.8 15.1W 36x160 47.0 36.01 0.650 12.000 1.020 5.9 50.0 9750 542 14.4 295 49 2.50 624.0 77.3 12.4W 36x150 44.2 35.85 0.625 11.975 0.940 6.4 52.0 9040 504 14.3 270 45 2.47 581.0 70.9 10.1W 36x135 39.7 35.55 0.600 11.950 0.790 7.6 54.1 7800 439 14.0 225 38 2.38 509.0 59.7 7.0W 33x619 181.0 38.47 1.970 16.910 3.540 2.4 15.2 41800 2170 15.2 2870 340 3.98 2560.0 537.0 567.0W 33x567 166.0 37.91 1.810 16.750 3.270 2.6 16.6 37700 1990 15.1 2580 308 3.94 2330.0 485.0 444.0W 33x515 151.0 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W 8x 67 19.7 9.00 0.570 8.280 0.935 4.4 11.1 272 60 3.7 89 21 2.12 70.2 32.7 5.1W 8x 58 17.1 8.75 0.510 8.220 0.810 5.1 12.4 228 52 3.7 75 18 2.10 59.8 27.9 3.3W 8x 48 14.1 8.50 0.400 8.110 0.685 5.9 15.8 184 43 3.6 61 15 2.08 49.0 22.9 2.0W 8x 40 11.7 8.25 0.360 8.070 0.560 7.2 17.6 146 36 3.5 49 12 2.04 39.8 18.5 1.1W 8x 35 10.3 8.12 0.310 8.020 0.495 8.1 20.4 127 31 3.5 43 11 2.03 34.7 16.1 0.8W 8x 31 9.1 8.00 0.285 7.995 0.435 9.2 22.2 110 28 3.5 37 9 2.02 30.4 14.1 0.5W 8x 28 8.2 8.06 0.285 6.535 0.465 7.0 22.2 98 24 3.5 22 7 1.62 27.2 10.1 0.5W 8x 24 7.1 7.93 0.245 6.495 0.400 8.1 25.8 83 21 3.4 18 6 1.61 23.2 8.6 0.3W 8x 21 6.2 8.28 0.250 5.270 0.400 6.6 27.5 75 18 3.5 10 4 1.26 20.4 5.7 0.3W 8x 18 5.3 8.14 0.230 5.250 0.330 8.0 29.9 62 15 3.4 8 3 1.23 17.0 4.7 0.2W 8x 15 4.4 8.11 0.245 4.015 0.315 6.4 28.1 48 12 3.3 3 2 0.88 13.6 2.7 0.1W 8x 13 3.8 7.99 0.230 4.000 0.255 7.8 29.9 40 10 3.2 3 1 0.84 11.4 2.2 0.1W 8x 10 3.0 7.89 0.170 3.940 0.205 9.6 40.5 31 8 3.2 2 1 0.84 8.9 1.7 0.0W 6x 25 7.3 6.38 0.320 6.080 0.455 6.7 15.5 53 17 2.7 17 6 1.52 18.9 8.6 0.5W 6x 20 5.9 6.20 0.260 6.020 0.365 8.2 19.1 41 13 2.7 13 4 1.50 14.9 6.7 0.2W 6x 15 4.4 5.99 0.230 5.990 0.260 11.5 21.6 29 10 2.6 9 3 1.46 10.8 4.8 0.1W 6x 16 4.7 6.28 0.260 4.030 0.405 5.0 19.1 32 10 2.6 4 2 0.97 11.7 3.4 0.2W 6x 12 3.5 6.03 0.230 4.000 0.280 7.1 21.6 22 7 2.5 3 2 0.92 8.3 2.3 0.1W 6x 9 2.7 5.90 0.170 3.940 0.215 9.2 29.2 16 6 2.5 2 1 0.90 6.2 1.7 0.0W 5x 19 5.5 5.15 0.270 5.030 0.430 5.8 14.0 26 10 2.2 9 4 1.28 11.6 5.5 0.3W 5x 16 4.7 5.01 0.240 5.000 0.360 6.9 15.8 21 9 2.1 8 3 1.27 9.6 4.6 0.2W 4x 13 3.8 4.16 0.280 4.060 0.345 5.9 10.6 11 5 1.7 4 2 1.00 6.3 2.9 0.2M 14x 18 5.1 14.00 0.215 4.000 0.270 7.4 60.3 148 21 5.4 3 1 0.72 24.9 2.2 0.1M 12x 12 3.5 12.00 0.177 3.065 0.225 6.8 62.5 72 12 4.6 1 1 0.53 14.3 1.1 0.1M 12x 11 3.2 11.97 0.160 3.065 0.210 7.3 63.6 65 11 4.6 1 1 0.54 13.2 1.0 0.0M 12x 10 2.9 11.97 0.149 3.250 0.180 9.1 68.0 62 10 4.6 1 1 0.58 12.2 1.0 0.0M 10x 9 2.7 10.00 0.157 2.690 0.206 6.5 58.4 39 8 3.8 1 0 0.48 9.2 0.8 0.0M 10x 8 2.3 9.95 0.141 2.690 0.182 7.4 59.3 34 7 3.8 1 0 0.43 8.2 0.7 0.0M 10x 8 2.2 9.99 0.130 2.690 0.173 7.8 65.0 33 7 3.8 0 0 0.47 7.7 0.6 0.0M 8x 6 1.9 8.00 0.135 2.281 0.189 6.0 53.8 18 5 3.1 0 0 0.42 5.4 0.5 0.0M 6x 4 1.3 6.00 0.114 1.844 0.171 5.4 47.0 7 2 2.4 0 0 0.36 2.8 0.3 0.0M 5x 19 5.6 5.00 0.316 5.003 0.416 6.0 11.2 24 10 2.1 8 3 1.19 11.0 5.0 0.3S 24x121 35.6 24.50 0.800 8.050 1.090 3.7 26.4 3160 258 9.4 83 21 1.53 306.0 36.2 12.8S 24x106 31.2 24.50 0.620 7.870 1.090 3.6 34.1 2940 240 9.7 77 20 1.57 279.0 33.2 10.1S 24x100 29.3 24.00 0.745 7.245 0.870 4.2 28.3 2390 199 9.0 48 13 1.27 240.0 23.9 7.6S 24x 90 26.5 24.00 0.625 7.125 0.870 4.1 33.7 2250 187 9.2 45 13 1.30 222.0 22.3 6.0S 24x 80 23.5 24.00 0.500 7.000 0.870 4.0 42.1 2100 175 9.5 42 12 1.34 204.0 20.7 4.9S 20x 96 28.2 20.30 0.800 7.200 0.920 3.9 21.6 1670 165 7.7 50 14 1.33 198.0 24.9 8.4S 20x 86 25.3 20.30 0.660 7.060 0.920 3.8 26.2 1580 155 7.9 47 13 1.36 183.0 23.0 6.6S 20x 75 22.0 20.00 0.635 6.385 0.795 4.0 27.1 1280 128 7.6 30 9 1.16 153.0 16.7 4.6S 20x 66 19.4 20.00 0.505 6.255 0.795 3.9 34.1 1190 119 7.8 28 9 1.19 140.0 15.3 3.6S 18x 70 20.6 18.00 0.711 6.251 0.691 4.5 21.8 926 103 6.7 24 8 1.08 125.0 14.4 4.2S 18x 55 16.1 18.00 0.461 6.001 0.691 4.3 33.6 804 89 7.1 21 7 1.14 105.0 12.1 2.4S 15x 50 14.7 15.00 0.550 5.640 0.622 4.5 23.2 486 65 5.8 16 6 1.03 77.1 10.0 2.1S 15x 43 12.6 15.00 0.411 5.501 0.622 4.4 31.0 447 60 5.9 14 5 1.07 69.3 9.0 1.5S 12x 50 14.7 12.00 0.687 5.477 0.659 4.2 13.9 305 51 4.6 16 6 1.03 61.2 10.3 2.8S 12x 41 12.0 12.00 0.462 5.252 0.659 4.0 20.7 272 45 4.8 14 5 1.06 53.1 8.9 1.8S 12x 35 10.3 12.00 0.428 5.078 0.544 4.7 23.4 229 38 4.7 10 4 0.98 44.8 6.8 1.1S 12x 32 9.4 12.00 0.350 5.000 0.544 4.6 28.6 218 36 4.8 9 4 1.00 42.0 6.4 0.9S 10x 35 10.3 10.00 0.594 4.944 0.491 5.0 13.8 147 29 3.8 8 3 0.90 35.4 6.2 1.3S 10x 25 7.5 10.00 0.311 4.661 0.491 4.7 26.4 124 25 4.1 7 3 0.95 28.4 5.0 0.6S 8x 23 6.8 8.00 0.441 4.171 0.426 4.9 14.5 65 16 3.1 4 2 0.80 19.3 3.7 0.6S 8x 18 5.4 8.00 0.271 4.001 0.426 4.7 23.7 58 14 3.3 4 2 0.83 16.5 3.2 0.3S 7x 20 5.9 7.00 0.450 3.860 0.392 4.9 12.3 42 12 2.7 3 2 0.73 14.5 3.0 0.4S 7x 15 4.5 7.00 0.252 3.662 0.392 4.7 21.9 37 10 2.9 3 1 0.77 12.1 2.4 0.2S 6x 17 5.1 6.00 0.465 3.565 0.359 5.0 9.9 26 9 2.3 2 1 0.68 10.6 2.4 0.4S 6x 12 3.7 6.00 0.232 3.332 0.359 4.6 19.9 22 7 2.5 2 1 0.70 8.5 1.9 0.2S 5x 15 4.3 5.00 0.494 3.284 0.326 5.0 7.5 15 6 1.9 2 1 0.62 7.4 1.9 0.3S 5x 10 2.9 5.00 0.214 3.004 0.326 4.6 17.4 12 5 2.0 1 1 0.64 5.7 1.4 0.1S 4x 10 2.8 4.00 0.326 2.796 0.293 4.8 8.7 7 3 1.6 1 1 0.57 4.0 1.1 0.1S 4x 8 2.3 4.00 0.193 2.663 0.293 4.5 14.7 6 3 1.6 1 1 0.58 3.5 1.0 0.1S 3x 8 2.2 3.00 0.349 2.509 0.260 4.8 5.6 3 2 1.1 1 0 0.52 2.4 0.8 0.1S 3x 6 1.7 3.00 0.170 2.330 0.260 4.5 11.4 3 2 1.2 0 0 0.52 2.0 0.7 0.0




bf2tf

hctw



C 15.x 50 14.7 15. 0.716 3.716 0.650 0 0 404.0 53.8 5.24 11. 3.78 0.87 68.20 8.17 2.67C 15.x 40 11.8 15. 0.520 3.520 0.650 0 0 349.0 46.5 5.44 9.23 3.37 0.89 57.20 6.87 1.46C 15.x 34 10.0 15. 0.400 3.400 0.650 0 0 315.0 42.0 5.62 8.13 3.11 0.90 50.40 6.23 1.02C 12.x 30 8.8 12. 0.510 3.170 0.501 0 0 162.0 27.0 4.29 5.14 2.06 0.76 33.60 4.33 0.87C 12.x 25 7.3 12. 0.387 3.047 0.501 0 0 144.0 24.1 4.43 4.47 1.88 0.78 29.20 3.84 0.54C 12.x 21 6.1 12. 0.282 2.942 0.501 0 0 129.0 21.5 4.61 3.88 1.73 0.80 25.40 3.49 0.37C 10.x 30 8.8 10. 0.673 3.033 0.436 0 0 103.0 20.7 3.42 3.94 1.65 0.67 26.60 3.78 1.23C 10.x 25 7.3 10. 0.526 2.886 0.436 0 0 91.2 18.2 3.52 3.36 1.48 0.68 23. 3.19 0.69C 10.x 20 5.9 10. 0.379 2.739 0.436 0 0 78.9 15.8 3.66 2.81 1.32 0.69 19.30 2.71 0.37C 10.x 15 4.5 10. 0.240 2.600 0.436 0 0 67.4 13.5 3.87 2.28 1.16 0.71 15.80 2.35 0.21C 9.x 20 5.9 9. 0.448 2.648 0.413 0 0 60.9 13.5 3.22 2.42 1.17 0.64 16.80 2.47 0.43C 9.x 15 4.4 9. 0.285 2.485 0.413 0 0 51.0 11.3 3.40 1.93 1.01 0.66 13.50 2.05 0.21C 9.x 13 3.9 9. 0.233 2.433 0.413 0 0 47.9 10.6 3.48 1.76 0.96 0.67 12.50 1.95 0.17C 8.x 19 5.5 8. 0.487 2.527 0.390 0 0 44.0 11.0 2.82 1.98 1.01 0.60 13.80 2.17 0.44C 8.x 14 4.0 8. 0.303 2.343 0.390 0 0 36.1 9.0 2.99 1.53 0.85 0.62 10.90 1.73 0.19C 8.x 12 3.4 8. 0.220 2.260 0.390 0 0 32.6 8.1 3.11 1.32 0.78 0.63 9.55 1.58 0.13C 7.x 15 4.3 7. 0.419 2.299 0.366 0 0 27.2 7.8 2.51 1.38 0.78 0.56 9.68 1.64 0.27C 7.x 12 3.6 7. 0.314 2.194 0.366 0 0 24.2 6.9 2.60 1.17 0.70 0.57 8.40 1.43 0.16C 7.x 10 2.9 7. 0.210 2.090 0.366 0 0 21.3 6.1 2.72 0.97 0.63 0.58 7.12 1.26 0.10C 6.x 13 3.8 6. 0.437 2.157 0.343 0 0 17.4 5.8 2.13 1.05 0.64 0.52 7.26 1.36 0.24C 6.x 11 3.1 6. 0.314 2.034 0.343 0 0 15.2 5.1 2.22 0.87 0.56 0.53 6.15 1.15 0.13C 6.x 8 2.4 6. 0.200 1.920 0.343 0 0 13.1 4.4 2.34 0.69 0.49 0.54 5.13 0.99 0.08C 5.x 9 2.6 5. 0.325 1.885 0.320 0 0 8.9 3.6 1.83 0.63 0.45 0.49 4.36 0.92 0.11C 5.x 7 2.0 5. 0.190 1.750 0.320 0 0 7.5 3.0 1.95 0.48 0.38 0.49 3.51 0.76 0.06C 4.x 7 2.1 4. 0.321 1.721 0.296 0 0 4.6 2.3 1.47 0.43 0.34 0.45 2.81 0.70 0.08C 4.x 5 1.6 4. 0.184 1.584 0.296 0 0 3.8 1.9 1.56 0.32 0.28 0.45 2.26 0.57 0.04C 3.x 6 1.8 3. 0.356 1.596 0.273 0 0 2.1 1.4 1.08 0.31 0.27 0.42 1.72 0.54 0.07C 3.x 5 1.5 3. 0.258 1.498 0.273 0 0 1.9 1.2 1.12 0.25 0.23 0.41 1.50 0.47 0.04C 3.x 4 1.2 3. 0.170 1.410 0.273 0 0 1.7 1.1 1.17 0.20 0.20 0.40 1.30 0.40 0.03

MC18.x 58 17.1 18. 0.700 4.200 0.625 0 0 676.0 75.1 6.29 17.80 5.32 1.02 94.60 9.94 2.81MC18.x 52 15.3 18. 0.600 4.100 0.625 0 0 627.0 69.7 6.41 16.40 5.07 1.04 86.50 9.13 2.03MC18.x 46 13.5 18. 0.500 4.000 0.625 0 0 578.0 64.3 6.56 15.10 4.82 1.06 78.40 8.42 1.45MC18.x 43 12.6 18. 0.450 3.950 0.625 0 0 554.0 61.6 6.64 14.40 4.69 1.07 74.40 8.10 1.23MC13.x 50 14.7 13. 0.787 4.412 0.610 0 0 314.0 48.4 4.62 16.50 4.79 1.06 60.50 10.10 2.98MC13.x 40 11.8 13. 0.560 4.185 0.610 0 0 273.0 42.0 4.82 13.70 4.26 1.08 50.90 8.57 1.57MC13.x 35 10.3 13. 0.447 4.072 0.610 0 0 252.0 38.8 4.95 12.30 3.99 1.10 46.20 7.95 1.14MC13.x 32 9.4 13. 0.375 4.000 0.610 0 0 239.0 36.8 5.06 11.40 3.81 1.11 43.10 7.60 0.94MC12.x 50 14.7 12. 0.835 4.135 0.700 0 0 269.0 44.9 4.28 17.40 5.65 1.09 56.10 10.20 3.24MC12.x 45 13.2 12. 0.712 4.012 0.700 0 0 252.0 42.0 4.36 15.80 5.33 1.09 51.70 9.35 2.35MC12.x 40 11.8 12. 0.590 3.890 0.700 0 0 234.0 39.0 4.46 14.30 5. 1.10 47.30 8.59 1.70MC12.x 35 10.3 12. 0.467 3.767 0.700 0 0 216.0 36.1 4.59 12.70 4.67 1.11 42.80 7.91 1.25MC12.x 31 9.1 12. 0.370 3.670 0.700 0 0 203.0 33.8 4.71 11.30 4.39 1.12 39.30 7.44 1.01MC12.x 11 3.1 12. 0.190 1.500 0.309 0 0 55.4 9.2 4.22 0.38 0.31 0.35 11.60 0.64 0.06MC10.x 41 12.1 10. 0.796 4.321 0.575 0 0 158.0 31.5 3.61 15.80 4.88 1.14 38.90 8.71 2.27MC10.x 34 9.9 10. 0.575 4.100 0.575 0 0 139.0 27.8 3.75 13.20 4.38 1.16 33.40 7.51 1.21MC10.x 29 8.4 10. 0.425 3.950 0.575 0 0 127.0 25.3 3.89 11.40 4.02 1.17 29.60 6.83 0.79MC10.x 25 7.3 10. 0.380 3.405 0.575 0 0 110.0 22.0 3.87 7.35 3. 1. 25.80 5.21 0.64MC10.x 22 6.4 10. 0.290 3.315 0.575 0 0 103.0 20.5 3.99 6.50 2.80 1. 23.60 4.86 0.51MC10.x 8 2.5 10. 0.170 1.500 0.280 0 0 32.0 6.4 3.61 0.33 0.27 0.37 7.86 0.55 0.04MC10.x 7 1.9 10. 0.152 1.127 0.202 0 0 22.1 4.4 3.40 0.11 0.12 0.24 5.73 0.25 0.02MC 9.x 25 7.5 9. 0.450 3.500 0.550 0 0 88.0 19.6 3.43 7.65 3.02 1.01 23.20 5.23 0.69MC 9.x 24 7.0 9. 0.400 3.450 0.550 0 0 85.0 18.9 3.48 7.22 2.93 1.01 22.20 5.05 0.60MC 8.x 23 6.7 8. 0.427 3.502 0.525 0 0 63.8 16.0 3.09 7.07 2.84 1.03 18.80 4.88 0.57MC 8.x 21 6.3 8. 0.375 3.450 0.525 0 0 61.6 15.4 3.13 6.64 2.74 1.03 18. 4.71 0.50MC 8.x 20 5.9 8. 0.400 3.025 0.500 0 0 54.5 13.6 3.05 4.47 2.05 0.87 16.20 3.57 0.44MC 8.x 19 5.5 8. 0.353 2.978 0.500 0 0 52.5 13.1 3.09 4.20 1.97 0.87 15.40 3.44 0.38MC 8.x 9 2.5 8. 0.179 1.874 0.311 0 0 23.3 5.8 3.05 0.63 0.43 0.50 6.91 0.88 0.06MC 7.x 23 6.7 7. 0.503 3.603 0.500 0 0 47.5 13.6 2.67 7.29 2.85 1.05 16.20 4.86 0.63MC 7.x 19 5.6 7. 0.352 3.452 0.500 0 0 43.2 12.3 2.77 6.11 2.57 1.04 14.30 4.34 0.41MC 6.x 18 5.3 6. 0.379 3.504 0.475 0 0 29.7 9.9 2.37 5.93 2.48 1.06 11.50 4.14 0.38MC 6.x 15 4.5 6. 0.340 3.500 0.385 0 0 25.4 8.5 2.38 4.97 2.03 1.05 9.82 3.28 0.22MC 6.x 16 4.8 6. 0.375 3.000 0.475 0 0 26.0 8.7 2.33 3.82 1.84 0.89 10.20 3.18 0.34MC 6.x 15 4.4 6. 0.316 2.941 0.475 0 0 25.0 8.3 2.37 3.51 1.75 0.89 9.69 3. 0.29MC 6.x 12 3.5 6. 0.310 2.497 0.375 0 0 18.7 6.2 2.30 1.87 1.04 0.73 7.38 1.79 0.15




bf2tf

hctw



WT18.x180. 52.7 18.70 1.120 16.730 2.010 0 14.1 1500.0 104.0 5.33 786.00 94.00 3.86 187.00 146.00 54.30WT18.x164. 48.2 18.54 1.020 16.630 1.850 0 15.4 1350.0 94.1 5.29 711.00 85.50 3.84 168.00 132.00 42.10WT18.x150. 44.1 18.37 0.945 16.655 1.680 0 16.7 1230.0 86.1 5.27 648.00 77.80 3.83 153.00 120.00 32.00WT18.x140. 41.2 18.26 0.885 16.595 1.570 0 17.8 1140.0 80.0 5.25 599.00 72.20 3.81 142.00 112.00 26.20WT18.x130. 38.2 18.13 0.840 16.550 1.440 0 18.7 1060.0 75.1 5.26 545.00 65.90 3.78 133.00 102.00 20.70WT18.x123. 36.0 18.04 0.800 16.510 1.350 0 19.7 995.0 71.0 5.26 507.00 61.40 3.75 125.00 94.90 17.30WT18.x115. 33.8 17.95 0.760 16.470 1.260 0 20.7 934.0 67.0 5.25 470.00 57.10 3.73 118.00 88.10 14.30WT18.x128. 37.7 18.71 0.960 12.215 1.730 0 16.9 1200.0 87.4 5.66 264.00 43.20 2.65 156.00 68.60 26.60WT18.x116. 34.1 18.56 0.870 12.120 1.570 0 18.7 1080.0 78.5 5.63 234.00 38.60 2.62 140.00 61.00 19.80WT18.x105. 30.9 18.34 0.830 12.180 1.360 0 19.6 985.0 73.1 5.65 206.00 33.80 2.58 131.00 53.50 13.90WT18.x 97. 28.5 18.25 0.765 12.115 1.260 0 21.2 901.0 67.0 5.62 187.00 30.90 2.56 120.00 48.90 11.10WT18.x 91. 26.8 18.17 0.725 12.075 1.180 0 22.4 845.0 63.1 5.62 174.00 28.80 2.55 113.00 45.40 9.19WT18.x 85. 25.0 18.08 0.680 12.030 1.100 0 23.9 786.0 58.9 5.61 160.00 26.60 2.53 105.00 41.90 7.51WT18.x 80. 23.5 18.00 0.650 12.000 1.020 0 25.0 740.0 55.8 5.61 147.00 24.60 2.50 100.00 38.60 6.17WT18.x 75. 22.1 17.92 0.625 11.975 0.940 0 26.0 698.0 53.1 5.62 135.00 22.50 2.47 95.50 35.50 5.04WT18.x 68. 19.9 17.77 0.600 11.950 0.790 0 27.1 637.0 49.7 5.66 113.00 18.90 2.38 94.30 29.80 3.48WT17.x177. 52.1 17.77 1.160 16.100 2.090 0 12.9 1320.0 96.8 5.03 729.00 90.60 3.74 174.00 141.00 57.20WT17.x159. 46.7 17.58 1.040 15.985 1.890 0 14.4 1160.0 85.8 4.99 645.00 80.70 3.71 154.00 125.00 42.10WT17.x146. 42.8 17.42 0.960 15.905 1.730 0 15.6 1050.0 78.3 4.97 581.00 73.10 3.69 140.00 113.00 32.40WT17.x132. 38.7 17.26 0.870 15.805 1.570 0 17.2 943.0 70.2 4.94 517.00 65.50 3.66 125.00 101.00 24.20WT17.x121. 35.4 17.09 0.830 15.860 1.400 0 18.1 871.0 65.8 4.96 466.00 58.80 3.63 116.00 90.90 17.90WT17.x111. 32.5 16.97 0.775 15.805 1.275 0 19.3 799.0 60.8 4.96 420.00 53.20 3.59 107.00 82.10 13.70WT17.x101. 29.5 16.84 0.715 15.745 1.150 0 21.0 725.0 55.5 4.95 375.00 47.60 3.56 97.70 73.40 10.20WT17.x 85. 24.8 16.91 0.670 11.500 1.220 0 22.4 649.0 51.1 5.12 155.00 27.00 2.50 90.80 42.20 8.80WT17.x 76. 22.4 16.75 0.635 11.565 1.055 0 23.6 592.0 47.4 5.14 136.00 23.60 2.47 84.50 37.00 6.16WT17.x 71. 20.8 16.65 0.605 11.535 0.960 0 24.8 552.0 44.7 5.15 123.00 21.30 2.43 79.80 33.50 4.84WT17.x 65. 19.2 16.55 0.580 11.510 0.855 0 25.8 513.0 42.1 5.18 109.00 18.90 2.39 75.60 29.70 3.67WT17.x 59. 17.3 16.43 0.550 11.480 0.740 0 27.3 469.0 39.2 5.20 93.60 16.30 2.32 74.80 25.70 2.64WT15.x118. 34.5 15.65 0.830 15.055 1.500 0 16.2 674.0 55.1 4.42 427.00 56.80 3.52 98.20 87.50 19.90WT15.x106. 31.0 15.47 0.775 15.105 1.315 0 17.4 610.0 50.5 4.43 378.00 50.10 3.49 89.50 77.20 13.90WT15.x 96. 28.1 15.34 0.710 15.040 1.185 0 19.0 549.0 45.7 4.42 336.00 44.70 3.46 80.80 68.90 10.30WT15.x 87. 25.4 15.22 0.655 14.985 1.065 0 20.6 497.0 41.7 4.42 299.00 39.90 3.43 73.40 61.40 7.61WT15.x 74. 21.7 15.34 0.650 10.480 1.180 0 20.8 466.0 40.6 4.63 113.00 21.70 2.28 72.20 34.00 7.30WT15.x 66. 19.4 15.15 0.615 10.545 1.000 0 22.0 421.0 37.4 4.66 98.00 18.60 2.25 66.80 29.20 4.85WT15.x 62. 18.2 15.09 0.585 10.515 0.930 0 23.1 396.0 35.3 4.66 90.40 17.20 2.23 63.10 27.00 3.98WT15.x 58. 17.1 15.01 0.565 10.495 0.850 0 23.9 373.0 33.7 4.67 82.10 15.70 2.19 60.40 24.60 3.21WT15.x 54. 15.9 14.91 0.545 10.475 0.760 0 24.8 349.0 32.0 4.69 73.00 13.90 2.15 57.70 22.00 2.49WT15.x 50. 14.5 14.82 0.520 10.450 0.670 0 26.0 322.0 30.0 4.71 63.90 12.20 2.10 57.40 19.30 1.88WT14.x109. 31.9 14.22 0.830 14.115 1.500 0 14.6 502.0 45.2 3.97 352.00 49.90 3.32 81.10 77.00 18.50WT14.x 97. 28.5 14.06 0.750 14.035 1.340 0 16.2 444.0 40.3 3.95 309.00 44.10 3.29 71.80 67.90 13.20WT14.x 89. 26.1 13.90 0.725 14.085 1.190 0 16.7 414.0 38.2 3.98 278.00 39.40 3.26 67.60 60.80 9.74WT14.x 81. 23.7 13.80 0.660 14.020 1.080 0 18.4 372.0 34.4 3.96 248.00 35.40 3.24 60.80 54.50 7.31WT14.x 73. 21.5 13.69 0.605 13.965 0.975 0 20.0 336.0 31.2 3.95 222.00 31.70 3.21 55.00 48.80 5.44WT14.x 65. 18.9 13.81 0.610 10.010 1.100 0 19.9 323.0 31.0 4.13 92.20 18.40 2.21 55.10 28.80 5.60WT14.x 57. 16.8 13.65 0.570 10.070 0.930 0 21.3 289.0 28.3 4.15 79.40 15.80 2.18 50.40 24.70 3.65WT14.x 51. 15.0 13.55 0.515 10.015 0.830 0 23.5 258.0 25.3 4.14 69.60 13.90 2.15 45.00 21.70 2.64WT14.x 47. 13.8 13.46 0.490 9.990 0.745 0 24.7 239.0 23.8 4.16 62.00 12.40 2.12 42.40 19.40 2.01WT14.x 42. 12.4 13.35 0.460 9.960 0.640 0 26.3 216.0 21.9 4.18 52.80 10.60 2.07 39.20 16.60 1.40WT12.x 88. 25.8 12.63 0.750 12.890 1.340 0 14.4 319.0 32.2 3.51 240.00 37.20 3.04 57.80 57.30 12.00WT12.x 81. 23.9 12.50 0.705 12.955 1.220 0 15.3 293.0 29.9 3.50 221.00 34.20 3.05 53.30 52.60 9.22WT12.x 73. 21.5 12.37 0.650 12.900 1.090 0 16.6 264.0 27.2 3.50 195.00 30.30 3.01 48.20 46.60 6.70WT12.x 66. 19.3 12.24 0.605 12.855 0.960 0 17.8 238.0 24.8 3.52 170.00 26.50 2.97 43.90 40.70 4.74WT12.x 59. 17.2 12.13 0.550 12.800 0.850 0 19.6 212.0 22.3 3.51 149.00 23.20 2.94 39.20 35.70 3.35WT12.x 52. 15.3 12.03 0.500 12.750 0.750 0 21.6 189.0 20.0 3.51 130.00 20.30 2.91 35.10 31.20 2.35WT12.x 52. 15.1 12.26 0.550 9.000 0.980 0 19.6 204.0 22.0 3.67 59.70 13.30 1.99 39.20 20.70 3.50WT12.x 47. 13.8 12.15 0.515 9.065 0.875 0 20.9 186.0 20.3 3.67 54.50 12.00 1.98 36.10 18.80 2.62WT12.x 42. 12.4 12.05 0.470 9.020 0.770 0 22.9 166.0 18.3 3.67 47.20 10.50 1.95 32.50 16.30 1.84WT12.x 38. 11.2 11.96 0.440 8.990 0.680 0 24.5 151.0 16.9 3.68 41.30 9.18 1.92 30.10 14.30 1.34WT12.x 34. 10.0 11.86 0.415 8.965 0.585 0 26.0 137.0 15.6 3.70 35.20 7.85 1.87 27.90 12.30 0.93WT12.x 31. 9.1 11.87 0.430 7.040 0.590 0 25.1 131.0 15.6 3.79 17.20 4.90 1.38 28.40 7.87 0.85WT12.x 28. 8.1 11.78 0.395 7.005 0.505 0 27.3 117.0 14.1 3.80 14.50 4.15 1.34 25.60 6.67 0.59WT11.x 83. 24.4 11.24 0.750 12.420 1.360 0 12.4 226.0 25.5 3.04 217.00 35.00 2.98 46.30 53.90 11.90WT11.x 74. 21.6 11.03 0.720 12.510 1.150 0 13.0 204.0 23.7 3.08 188.00 30.00 2.95 42.40 46.30 7.69WT11.x 66. 19.4 10.91 0.650 12.440 1.035 0 14.4 181.0 21.1 3.06 166.00 26.70 2.93 37.60 41.10 5.62WT11.x 61. 17.9 10.84 0.600 12.390 0.960 0 15.6 166.0 19.3 3.04 152.00 24.60 2.92 34.30 37.80 4.47WT11.x 56. 16.3 10.76 0.550 12.340 0.875 0 17.1 150.0 17.5 3.03 137.00 22.20 2.90 31.00 34.10 3.40WT11.x 51. 14.9 10.68 0.500 12.290 0.800 0 18.8 135.0 15.8 3.01 124.00 20.20 2.89 27.90 30.90 2.60WT11.x 47. 13.7 10.81 0.580 8.420 0.930 0 16.2 144.0 17.9 3.25 46.40 11.00 1.84 31.80 17.40 3.01WT11.x 42. 12.2 10.72 0.515 8.355 0.835 0 18.2 127.0 15.7 3.22 40.70 9.75 1.83 28.00 15.30 2.16WT11.x 37. 10.7 10.62 0.455 8.295 0.740 0 20.6 110.0 13.8 3.21 35.30 8.51 1.81 24.40 13.30 1.51WT11.x 34. 10.0 10.56 0.430 8.270 0.685 0 21.8 103.0 12.9 3.20 32.40 7.83 1.80 22.90 12.20 1.22WT11.x 31. 9.1 10.49 0.400 8.240 0.615 0 23.5 93.8 11.9 3.21 28.70 6.97 1.77 21.10 10.90 0.51WT11.x 29. 8.4 10.53 0.405 6.555 0.650 0 23.2 90.4 11.8 3.29 15.30 4.67 1.35 21.20 7.42 0.88WT11.x 25. 7.4 10.41 0.380 6.530 0.535 0 24.7 80.3 10.7 3.30 12.50 3.82 1.30 20.80 6.09 0.57WT11.x 22. 6.5 10.33 0.350 6.500 0.450 0 26.8 71.1 9.7 3.31 10.30 3.18 1.26 17.60 5.09 0.38WT 9.x 72. 21.0 9.74 0.730 11.220 1.320 0 11.0 142.0 18.5 2.60 156.00 27.70 2.72 34.00 42.70 9.70WT 9.x 65. 19.1 9.63 0.670 11.160 1.200 0 11.9 127.0 16.7 2.58 139.00 24.90 2.70 30.50 38.30 7.30WT 9.x 60. 17.5 9.48 0.655 11.265 1.060 0 12.3 119.0 15.9 2.60 126.00 22.50 2.69 28.70 34.60 5.30WT 9.x 53. 15.6 9.36 0.590 11.200 0.940 0 13.6 104.0 14.1 2.59 110.00 19.70 2.66 25.20 30.20 3.73WT 9.x 49. 14.3 9.30 0.535 11.145 0.870 0 15.0 93.8 12.7 2.56 100.00 18.00 2.65 22.60 27.60 2.92WT 9.x 43. 12.7 9.19 0.480 11.090 0.770 0 16.7 82.4 11.2 2.55 87.60 15.80 2.63 19.90 24.20 2.04WT 9.x 38. 11.2 9.10 0.425 11.035 0.680 0 18.9 71.8 9.8 2.54 76.20 13.80 2.61 17.30 21.10 1.41WT 9.x 36. 10.4 9.23 0.495 7.635 0.810 0 16.2 78.2 11.2 2.74 30.10 7.89 1.70 20.00 12.30 1.74WT 9.x 33. 9.6 9.18 0.450 7.590 0.750 0 17.8 70.7 10.1 2.72 27.40 7.22 1.69 18.00 11.20 1.36WT 9.x 30. 8.8 9.12 0.415 7.555 0.695 0 19.3 64.7 9.3 2.71 25.00 6.63 1.69 16.50 10.30 1.08WT 9.x 28. 8.1 9.06 0.390 7.530 0.630 0 20.6 59.5 8.6 2.71 22.50 5.97 1.67 15.30 9.27 0.83WT 9.x 25. 7.3 8.99 0.355 7.495 0.570 0 22.6 53.5 7.8 2.70 20.00 5.35 1.65 13.80 8.29 0.61WT 9.x 23. 6.8 9.03 0.360 6.060 0.605 0 22.3 52.1 7.8 2.77 11.30 3.72 1.29 13.90 5.85 0.61WT 9.x 20. 5.9 8.95 0.315 6.015 0.525 0 25.5 44.8 6.7 2.76 9.55 3.17 1.27 12.00 4.97 0.40WT 9.x 18. 5.2 8.85 0.300 6.000 0.425 0 26.8 40.1 6.2 2.79 7.67 2.56 1.22 12.00 4.03 0.25




bf2tf

hctw



WT 8.x 50. 14.7 8.48 0.585 10.425 0.985 0 12.1 76.8 11.4 2.28 93.10 17.90 2.51 20.70 27.40 3.85WT 8.x 45. 13.1 8.38 0.525 10.365 0.875 0 13.5 67.2 10.1 2.27 81.30 15.70 2.49 18.10 24.00 2.72WT 8.x 39. 11.3 8.26 0.455 10.295 0.760 0 15.6 56.9 8.6 2.24 69.20 13.40 2.47 15.30 20.50 1.78WT 8.x 34. 9.8 8.16 0.395 10.235 0.665 0 18.0 48.6 7.4 2.22 59.50 11.60 2.46 13.00 17.70 1.19WT 8.x 29. 8.4 8.22 0.430 7.120 0.715 0 16.5 48.7 7.8 2.41 21.60 6.06 1.60 13.80 9.43 1.10WT 8.x 25. 7.4 8.13 0.380 7.070 0.630 0 18.7 42.3 6.8 2.40 18.60 5.26 1.59 12.00 8.16 0.76WT 8.x 23. 6.6 8.06 0.345 7.035 0.565 0 20.6 37.8 6.1 2.39 16.40 4.67 1.57 10.80 7.23 0.65WT 8.x 20. 5.9 8.01 0.305 6.995 0.505 0 23.3 33.1 5.3 2.37 14.40 4.12 1.57 9.43 6.37 0.40WT 8.x 18. 5.3 7.93 0.295 6.985 0.430 0 24.1 30.6 5.1 2.41 12.20 3.50 1.52 8.93 5.42 0.27WT 8.x 16. 4.6 7.94 0.275 5.525 0.440 0 25.8 27.4 4.6 2.45 6.20 2.24 1.17 8.27 3.52 0.23WT 8.x 13. 3.8 7.84 0.250 5.500 0.345 0 28.4 23.5 4.1 2.47 4.80 1.74 1.12 8.12 2.74 0.13WT 7.x365. 107.0 11.21 3.070 17.890 4.910 0 1.9 739.0 95.4 2.62 2360.00 264.00 4.69 211.00 408.00 714.00WT 7.x333. 97.8 10.82 2.830 17.650 4.520 0 2.0 622.0 82.1 2.52 2080.00 236.00 4.62 182.00 365.00 555.00WT 7.x303. 88.9 10.46 2.595 17.415 4.160 0 2.2 524.0 70.6 2.43 1840.00 211.00 4.55 157.00 326.00 430.00WT 7.x275. 80.9 10.12 2.380 17.200 3.820 0 2.4 442.0 60.9 2.34 1630.00 189.00 4.49 136.00 292.00 331.00WT 7.x250. 73.5 9.80 2.190 17.010 3.500 0 2.6 375.0 52.7 2.26 1440.00 169.00 4.43 117.00 261.00 255.00WT 7.x228. 66.9 9.51 2.015 16.835 3.210 0 2.8 321.0 45.9 2.19 1280.00 152.00 4.38 102.00 234.00 196.00WT 7.x213. 62.6 9.34 1.875 16.695 3.035 0 3.0 287.0 41.4 2.14 1180.00 141.00 4.34 91.70 217.00 164.00WT 7.x199. 58.5 9.15 1.770 16.590 2.845 0 3.2 257.0 37.6 2.10 1090.00 131.00 4.31 82.90 201.00 135.00WT 7.x185. 54.4 8.96 1.655 16.475 2.660 0 3.4 229.0 33.9 2.05 994.00 121.00 4.27 74.40 185.00 110.00WT 7.x171. 50.3 8.77 1.540 16.360 2.470 0 3.7 203.0 30.4 2.01 903.00 110.00 4.24 66.20 169.00 88.30WT 7.x156. 45.7 8.56 1.410 16.230 2.260 0 4.0 176.0 26.7 1.96 807.00 99.40 4.20 57.70 152.00 67.50WT 7.x142. 41.6 8.37 1.290 16.110 2.070 0 4.4 153.0 23.5 1.92 722.00 89.70 4.17 50.40 137.00 51.80WT 7.x129. 37.8 8.19 1.175 15.995 1.890 0 4.9 133.0 20.7 1.88 645.00 80.70 4.13 43.90 123.00 39.30WT 7.x117. 34.2 8.02 1.070 15.890 1.720 0 5.3 116.0 18.2 1.84 576.00 72.50 4.10 38.20 110.00 29.60WT 7.x106. 31.0 7.86 0.980 15.800 1.560 0 5.8 102.0 16.2 1.81 513.00 65.00 4.07 33.40 99.00 22.20WT 7.x 97. 28.4 7.74 0.890 15.710 1.440 0 6.4 89.8 14.4 1.78 466.00 59.30 4.05 29.40 90.20 17.30WT 7.x 88. 25.9 7.61 0.830 15.650 1.310 0 6.9 80.5 13.0 1.76 419.00 53.50 4.02 26.30 81.40 13.20WT 7.x 80. 23.4 7.49 0.745 15.565 1.190 0 7.7 70.2 11.4 1.73 374.00 48.10 4.00 22.80 73.00 9.84WT 7.x 73. 21.3 7.39 0.680 15.500 1.090 0 8.4 62.5 10.2 1.71 338.00 43.70 3.98 20.20 66.30 7.56WT 7.x 66. 19.4 7.33 0.645 14.725 1.030 0 8.8 57.8 9.6 1.73 274.00 37.20 3.76 18.60 56.60 6.13WT 7.x 60. 17.7 7.24 0.590 14.670 0.940 0 9.7 51.7 8.6 1.71 247.00 33.70 3.74 16.50 51.20 4.67WT 7.x 55. 16.0 7.16 0.525 14.605 0.860 0 10.9 45.3 7.6 1.68 223.00 30.60 3.73 14.40 46.40 3.55WT 7.x 50. 14.6 7.08 0.485 14.565 0.780 0 11.8 40.9 6.9 1.67 201.00 27.60 3.71 12.90 41.80 2.68WT 7.x 45. 13.2 7.01 0.440 14.520 0.710 0 13.0 36.4 6.2 1.66 181.00 25.00 3.70 11.50 37.80 2.03WT 7.x 41. 12.0 7.16 0.510 10.130 0.855 0 11.2 41.2 7.1 1.85 74.20 14.60 2.48 13.20 22.40 2.53WT 7.x 37. 10.9 7.09 0.450 10.070 0.785 0 12.7 36.0 6.3 1.82 66.90 13.30 2.48 11.50 20.30 1.94WT 7.x 34. 10.0 7.02 0.415 10.035 0.720 0 13.7 32.6 5.7 1.81 60.70 12.10 2.46 10.40 18.50 1.51WT 7.x 31. 9.0 6.95 0.375 9.995 0.645 0 15.2 28.9 5.1 1.80 53.70 10.70 2.45 9.16 16.40 1.10WT 7.x 27. 7.8 6.96 0.370 8.060 0.660 0 15.4 27.6 4.9 1.88 28.80 7.16 1.92 8.87 11.00 0.97WT 7.x 24. 7.1 6.89 0.340 8.030 0.595 0 16.8 24.9 4.5 1.87 25.70 6.40 1.91 8.00 9.82 0.73WT 7.x 22. 6.3 6.83 0.305 7.995 0.530 0 18.7 21.9 4.0 1.86 22.60 5.65 1.89 7.05 8.66 0.52WT 7.x 19. 5.6 7.05 0.310 6.770 0.515 0 19.8 23.3 4.2 2.04 13.30 3.94 1.55 7.45 6.07 0.40WT 7.x 17. 5.0 6.99 0.285 6.745 0.455 0 21.5 20.9 3.8 2.04 11.70 3.45 1.53 6.74 5.32 0.28WT 7.x 15. 4.4 6.92 0.270 6.730 0.385 0 22.7 19.0 3.5 2.07 9.79 2.91 1.49 6.25 4.49 0.19WT 7.x 13. 3.8 6.95 0.255 5.025 0.420 0 24.1 17.3 3.3 2.12 4.45 1.77 1.08 5.89 2.77 0.18WT 7.x 11. 3.3 6.87 0.230 5.000 0.335 0 26.7 14.8 2.9 2.14 3.50 1.40 1.04 5.20 2.19 0.10WT 6.x168. 49.4 8.41 1.775 13.385 2.955 0 2.7 190.0 31.2 1.96 593.00 88.60 3.47 68.40 137.00 120.00WT 6.x153. 44.8 8.16 1.625 13.235 2.705 0 3.0 162.0 27.0 1.90 525.00 79.30 3.42 59.10 122.00 92.00WT 6.x140. 41.0 7.93 1.530 13.140 2.470 0 3.2 141.0 24.1 1.86 469.00 71.30 3.38 51.90 110.00 70.90WT 6.x126. 37.0 7.70 1.395 13.005 2.250 0 3.5 121.0 20.9 1.81 414.00 63.60 3.34 44.80 97.90 53.50WT 6.x115. 33.9 7.53 1.285 12.895 2.070 0 3.8 106.0 18.5 1.77 371.00 57.50 3.31 39.40 88.40 41.60WT 6.x105. 30.9 7.36 1.180 12.790 1.900 0 4.1 92.1 16.4 1.73 332.00 51.90 3.28 34.50 79.70 32.20WT 6.x 95. 27.9 7.19 1.060 12.670 1.735 0 4.6 79.0 14.2 1.68 295.00 46.50 3.25 29.80 71.30 24.40WT 6.x 85. 25.0 7.01 0.960 12.570 1.560 0 5.1 67.8 12.3 1.65 259.00 41.20 3.22 25.60 63.00 17.70WT 6.x 76. 22.4 6.86 0.870 12.480 1.400 0 5.6 58.5 10.8 1.62 227.00 36.40 3.19 22.00 55.60 12.80WT 6.x 68. 20.0 6.70 0.790 12.400 1.250 0 6.1 50.6 9.5 1.59 199.00 32.10 3.16 19.00 49.00 9.22WT 6.x 60. 17.6 6.56 0.710 12.320 1.105 0 6.8 43.4 8.2 1.57 172.00 28.00 3.13 16.20 42.70 6.43WT 6.x 53. 15.6 6.45 0.610 12.220 0.990 0 8.0 36.3 6.9 1.53 151.00 24.70 3.11 13.60 37.50 4.55WT 6.x 48. 14.1 6.36 0.550 12.160 0.900 0 8.8 32.0 6.1 1.51 135.00 22.20 3.09 11.90 33.70 3.42WT 6.x 44. 12.8 6.26 0.515 12.125 0.810 0 9.4 28.9 5.6 1.50 120.00 19.90 3.07 10.70 30.20 2.54WT 6.x 40. 11.6 6.19 0.470 12.080 0.735 0 10.3 25.8 5.0 1.49 108.00 17.90 3.05 9.49 27.20 1.92WT 6.x 36. 10.6 6.13 0.430 12.040 0.670 0 11.3 23.2 4.5 1.48 97.50 16.20 3.04 8.48 24.60 1.46WT 6.x 33. 9.5 6.06 0.390 12.000 0.605 0 12.4 20.6 4.1 1.47 87.20 14.50 3.02 7.50 22.00 1.09WT 6.x 29. 8.5 6.09 0.360 10.010 0.640 0 13.5 19.1 3.8 1.50 53.50 10.70 2.51 6.97 16.30 1.05WT 6.x 27. 7.8 6.03 0.345 9.995 0.575 0 14.1 17.7 3.5 1.51 47.90 9.58 2.48 6.46 14.60 0.79WT 6.x 25. 7.3 6.09 0.370 8.080 0.640 0 13.1 18.7 3.8 1.60 28.20 6.97 1.96 6.90 10.70 0.89WT 6.x 23. 6.6 6.03 0.335 8.045 0.575 0 14.5 16.6 3.4 1.58 25.00 6.21 1.94 6.12 9.50 0.66WT 6.x 20. 5.9 5.97 0.295 8.005 0.515 0 16.5 14.4 3.0 1.57 22.00 5.51 1.93 5.30 8.41 0.48WT 6.x 18. 5.2 6.25 0.300 6.560 0.520 0 18.1 16.0 3.2 1.76 12.20 3.73 1.54 5.71 5.73 0.37WT 6.x 15. 4.4 6.17 0.260 6.520 0.440 0 20.9 13.5 2.8 1.75 10.20 3.12 1.52 4.83 4.78 0.23WT 6.x 13. 3.8 6.11 0.230 6.490 0.380 0 23.6 11.7 2.4 1.75 8.66 2.67 1.51 4.20 4.08 0.15WT 6.x 11. 3.2 6.16 0.260 4.030 0.425 0 20.9 11.7 2.6 1.90 2.33 1.16 0.85 4.63 1.83 0.15WT 6.x 10. 2.8 6.08 0.235 4.005 0.350 0 23.1 10.1 2.3 1.90 1.88 0.94 0.82 4.11 1.49 0.09WT 6.x 8. 2.4 5.99 0.220 3.990 0.265 0 24.7 8.7 2.0 1.92 1.41 0.71 0.77 3.72 1.13 0.05WT 6.x 7. 2.1 5.95 0.200 3.970 0.225 0 27.2 7.7 1.8 1.92 1.18 0.59 0.75 3.32 0.95 0.04WT 5.x 56. 16.5 5.68 0.755 10.415 1.250 0 5.2 28.6 6.4 1.32 118.00 22.60 2.68 13.40 34.60 7.50WT 5.x 50. 14.7 5.55 0.680 10.340 1.120 0 5.8 24.5 5.6 1.29 103.00 20.00 2.65 11.40 30.50 5.41WT 5.x 44. 12.9 5.42 0.605 10.265 0.990 0 6.5 20.8 4.8 1.27 89.30 17.40 2.63 9.65 26.50 3.75WT 5.x 39. 11.3 5.30 0.530 10.190 0.870 0 7.4 17.4 4.0 1.24 76.80 15.10 2.60 8.06 22.90 2.55WT 5.x 34. 10.0 5.20 0.470 10.130 0.770 0 8.4 14.9 3.5 1.22 66.80 13.20 2.59 6.85 20.00 1.78WT 5.x 30. 8.8 5.11 0.420 10.080 0.680 0 9.4 12.9 3.0 1.21 58.10 11.50 2.57 5.87 17.50 1.23WT 5.x 27. 7.9 5.05 0.370 10.030 0.615 0 10.6 11.1 2.6 1.19 51.70 10.30 2.56 5.05 15.70 0.91WT 5.x 25. 7.2 4.99 0.340 10.000 0.560 0 11.6 10.0 2.4 1.18 46.70 9.34 2.54 4.52 14.20 0.69WT 5.x 23. 6.6 5.05 0.350 8.020 0.620 0 11.2 10.2 2.5 1.24 26.70 6.65 2.01 4.65 10.10 0.75WT 5.x 20. 5.7 4.96 0.315 7.985 0.530 0 12.5 8.8 2.2 1.24 22.50 5.64 1.98 3.99 8.59 0.49WT 5.x 17. 4.8 4.86 0.290 7.960 0.435 0 13.6 7.7 1.9 1.26 18.30 4.60 1.94 3.48 7.01 0.29WT 5.x 15. 4.4 5.24 0.300 5.810 0.510 0 14.8 9.3 2.2 1.45 8.35 2.87 1.37 4.01 4.42 0.31WT 5.x 13. 3.8 5.16 0.260 5.770 0.440 0 17.0 7.9 1.9 1.44 7.05 2.44 1.36 3.39 3.75 0.20WT 5.x 11. 3.2 5.09 0.240 5.750 0.360 0 18.4 6.9 1.7 1.46 5.71 1.99 1.33 3.02 3.05 0.12WT 5.x 10. 2.8 5.12 0.250 4.020 0.395 0 17.7 6.7 1.7 1.54 2.15 1.07 0.87 3.10 1.68 0.12WT 5.x 9. 2.5 5.05 0.240 4.010 0.330 0 18.4 6.1 1.6 1.56 1.78 0.89 0.84 2.90 1.40 0.08WT 5.x 8. 2.2 4.99 0.230 4.000 0.270 0 19.2 5.4 1.5 1.57 1.45 0.72 0.81 3.03 1.15 0.05WT 5.x 6. 1.8 4.93 0.190 3.960 0.210 0 23.3 4.3 1.2 1.57 1.09 0.55 0.79 2.50 0.87 0.03




bf2tf

hctw



WT 4.x 34. 9.8 4.50 0.570 8.280 0.935 0 5.6 10.9 3.0 1.05 44.30 10.70 2.12 6.29 16.30 2.52WT 4.x 29. 8.6 4.38 0.510 8.220 0.810 0 6.2 9.1 2.6 1.03 37.50 9.13 2.10 5.25 13.90 1.66WT 4.x 24. 7.1 4.25 0.400 8.110 0.685 0 7.9 6.8 2.0 0.99 30.50 7.52 2.08 3.94 11.40 0.98WT 4.x 20. 5.9 4.13 0.360 8.070 0.560 0 8.8 5.7 1.7 0.99 24.50 6.08 2.04 3.25 9.25 0.56WT 4.x 18. 5.1 4.06 0.310 8.020 0.495 0 10.2 4.8 1.4 0.97 21.30 5.31 2.03 2.71 8.06 0.38WT 4.x 16. 4.6 4.00 0.285 7.995 0.435 0 11.1 4.3 1.3 0.97 18.50 4.64 2.02 2.39 7.04 0.27WT 4.x 14. 4.1 4.03 0.285 6.535 0.465 0 11.1 4.2 1.3 1.01 10.80 3.31 1.62 2.38 5.05 0.27WT 4.x 12. 3.5 3.96 0.245 6.495 0.400 0 12.9 3.5 1.1 1.00 9.14 2.81 1.61 1.98 4.29 0.17WT 4.x 11. 3.1 4.14 0.250 5.270 0.400 0 13.8 3.9 1.2 1.12 4.89 1.85 1.26 2.11 2.84 0.14WT 4.x 9. 2.6 4.07 0.230 5.250 0.330 0 15.0 3.4 1.0 1.14 3.98 1.52 1.23 1.86 2.33 0.09WT 4.x 8. 2.2 4.05 0.245 4.015 0.315 0 14.0 3.3 1.1 1.22 1.70 0.85 0.88 1.91 1.33 0.07WT 4.x 7. 1.9 3.99 0.230 4.000 0.255 0 15.0 2.9 1.0 1.23 1.37 0.68 0.84 1.74 1.08 0.04WT 4.x 5. 1.5 3.94 0.170 3.940 0.205 0 20.2 2.2 0.7 1.20 1.05 0.53 0.84 1.27 0.83 0.02WT 3.x 13. 3.7 3.19 0.320 6.080 0.455 0 7.8 2.3 0.9 0.79 8.53 2.81 1.52 1.68 4.28 0.23WT 3.x 10. 2.9 3.10 0.260 6.020 0.365 0 9.6 1.8 0.7 0.77 6.64 2.21 1.50 1.29 3.36 0.12WT 3.x 8. 2.2 2.99 0.230 5.990 0.260 0 10.8 1.4 0.6 0.80 4.66 1.56 1.45 1.03 2.37 0.05WT 3.x 8. 2.4 3.14 0.260 4.030 0.405 0 9.6 1.7 0.7 0.84 2.21 1.10 0.97 1.25 1.70 0.11WT 3.x 6. 1.8 3.02 0.230 4.000 0.280 0 10.8 1.3 0.6 0.86 1.50 0.75 0.92 1.01 1.16 0.05WT 3.x 5. 1.3 2.95 0.170 3.940 0.215 0 14.6 0.9 0.4 0.84 1.10 0.56 0.90 0.72 0.86 0.02WT 3.x 10. 2.8 2.58 0.270 5.030 0.430 0 7.0 1.0 0.5 0.61 4.56 1.82 1.28 0.97 2.76 0.15WT 3.x 8. 2.3 2.51 0.240 5.000 0.360 0 7.9 0.9 0.4 0.60 3.75 1.50 1.27 0.80 2.29 0.09WT 2.x 7. 1.9 2.08 0.280 4.060 0.345 0 5.3 0.5 0.3 0.52 1.93 0.95 1.00 0.62 1.46 0.08

MT 7.x 9. 2.5 7.00 0.215 4.000 0.270 0 30.1 13.1 2.7 2.27 1.32 0.66 0.72 4.88 1.16 0.06MT 6.x 6. 1.7 6.00 0.177 3.065 0.225 0 31.3 6.6 1.6 1.95 0.49 0.32 0.53 2.89 0.58 0.03MT 5.x 5. 1.3 5.00 0.157 2.690 0.206 0 29.2 3.5 1.0 1.62 0.31 0.23 0.48 1.81 0.41 0.02MT 4.x 3. 1.0 4.00 0.135 2.281 0.189 0 26.9 1.6 0.6 1.28 0.17 0.15 0.42 1.01 0.26 0.01MT 3.x 2. 0.6 3.00 0.114 1.844 0.171 0 23.5 0.6 0.3 0.94 0.08 0.09 0.36 0.52 0.16 0.01MT 3.x 9. 2.8 2.50 0.316 5.003 0.416 0 5.6 1.0 0.5 0.62 3.93 1.57 1.19 1.03 2.66 0.17

ST12.x 61. 17.8 12.25 0.800 8.050 1.090 0 13.2 259.0 30.1 3.82 41.70 10.40 1.53 54.50 18.10 6.38ST12.x 53. 15.6 12.25 0.620 7.870 1.090 0 17.0 216.0 24.1 3.72 38.50 9.80 1.57 43.30 16.60 5.04ST12.x 50. 14.7 12.00 0.745 7.245 0.870 0 14.1 215.0 26.3 3.83 23.80 6.58 1.27 47.50 12.00 3.76ST12.x 45. 13.2 12.00 0.625 7.125 0.870 0 16.8 190.0 22.6 3.79 22.50 6.31 1.30 41.10 11.20 3.01ST12.x 40. 11.7 12.00 0.500 7.000 0.870 0 21.1 162.0 18.7 3.72 21.10 6.04 1.34 33.60 10.40 2.43ST10.x 48. 14.1 10.15 0.800 7.200 0.920 0 10.8 143.0 20.3 3.18 25.10 6.97 1.33 36.90 12.50 4.15ST10.x 43. 12.7 10.15 0.660 7.060 0.920 0 13.1 125.0 17.2 3.14 23.40 6.63 1.36 31.10 11.60 3.30ST10.x 38. 11.0 10.00 0.635 6.385 0.795 0 13.6 109.0 15.8 3.15 14.90 4.66 1.16 28.60 8.37 2.28ST10.x 33. 9.7 10.00 0.505 6.225 0.795 0 17.0 93.1 12.9 3.10 13.80 4.43 1.19 23.40 7.70 1.78ST 9.x 35. 10.3 9.00 0.711 6.251 0.691 0 10.9 84.7 14.0 2.87 12.10 3.86 1.08 25.10 7.21 2.05ST 9.x 27. 8.0 9.00 0.461 6.001 0.691 0 16.8 62.4 9.6 2.79 10.40 3.47 1.14 17.30 6.07 1.18ST 8.x 25. 7.3 7.50 0.550 5.640 0.622 0 11.6 40.6 7.7 2.35 7.85 2.78 1.03 14.00 5.01 1.05ST 8.x 21. 6.3 7.50 0.411 5.501 0.622 0 15.5 33.0 6.0 2.29 7.19 2.61 1.07 10.80 4.54 0.77ST 6.x 25. 7.3 6.00 0.687 5.477 0.659 0 7.0 25.2 6.1 1.85 7.85 2.87 1.03 11.00 5.19 1.39ST 6.x 20. 6.0 6.00 0.462 5.252 0.659 0 10.3 18.9 4.3 1.78 6.78 2.58 1.06 7.71 4.45 0.87ST 6.x 18. 5.2 6.00 0.428 5.078 0.545 0 11.7 17.2 4.0 1.83 4.94 1.95 0.98 7.12 3.41 0.54ST 6.x 16. 4.7 6.00 0.350 5.000 0.544 0 14.3 14.9 3.3 1.78 4.68 1.87 1.00 5.94 3.22 0.45ST 5.x 18. 5.2 5.00 0.594 4.944 0.491 0 6.9 12.5 3.6 1.56 4.18 1.69 0.90 6.58 3.11 0.63ST 5.x 13. 3.7 5.00 0.311 4.661 0.491 0 13.2 7.8 2.1 1.45 3.39 1.46 0.95 3.70 2.49 0.30ST 4.x 12. 3.4 4.00 0.441 4.171 0.425 0 7.3 5.0 1.8 1.22 2.15 1.03 0.80 3.19 1.84 0.27ST 4.x 9. 2.7 4.00 0.271 4.001 0.425 0 11.8 3.5 1.1 1.14 1.86 0.93 0.83 2.07 1.59 0.17ST 4.x 10. 2.9 3.50 0.450 3.860 0.392 0 6.1 3.4 1.4 1.07 1.59 0.82 0.73 2.47 1.48 0.22ST 4.x 8. 2.3 3.50 0.252 3.662 0.392 0 10.9 2.2 0.8 0.99 1.32 0.72 0.77 1.48 1.23 0.12ST 3.x 9. 2.5 3.00 0.465 3.565 0.359 0 5.0 2.1 1.0 0.92 1.15 0.65 0.68 1.85 1.18 0.18ST 3.x 6. 1.8 3.00 0.232 3.332 0.359 0 10.0 1.3 0.6 0.83 0.91 0.55 0.70 1.01 0.93 0.08ST 3.x 7. 2.2 2.50 0.494 3.284 0.326 0 3.8 1.3 0.7 0.76 0.83 0.51 0.62 1.34 0.94 0.16ST 3.x 5. 1.5 2.50 0.214 3.004 0.326 0 8.7 0.7 0.4 0.68 0.61 0.41 0.64 0.65 0.69 0.06ST 2.x 5. 1.4 2.00 0.326 2.796 0.293 0 4.3 0.5 0.3 0.58 0.45 0.32 0.57 0.59 0.57 0.06ST 2.x 4. 1.1 2.00 0.193 2.663 0.293 0 7.3 0.3 0.2 0.53 0.38 0.29 0.58 0.38 0.48 0.04ST 2.x 4. 1.1 1.50 0.349 2.509 0.260 0 2.8 0.2 0.2 0.43 0.29 0.23 0.52 0.35 0.41 0.04ST 2.x 3. 0.8 1.50 0.170 2.330 0.260 0 5.7 0.1 0.1 0.38 0.23 0.19 0.52 0.20 0.33 0.02

HP14.x117. 34.4 14.21 0.805 14.885 0.805 9.2 14.2 1220.0 172.0 5.96 443.00 59.50 3.59 194.00 91.40 8.02HP14.x102. 30.0 14.01 0.705 14.785 0.705 10.5 16.2 1050.0 150.0 5.92 380.00 51.40 3.56 169.00 78.80 5.40HP14.x 89. 26.1 13.83 0.615 14.695 0.615 11.9 18.5 904.0 131.0 5.88 326.00 44.30 3.53 146.00 67.70 3.60HP14.x 73. 21.4 13.61 0.505 14.585 0.505 14.4 22.6 729.0 107.0 5.84 261.00 35.80 3.49 118.00 54.60 2.01HP13.x100. 29.4 13.15 0.765 13.205 0.765 8.6 13.6 886.0 135.0 5.49 294.00 44.50 3.16 153.00 68.60 6.25HP13.x 87. 25.5 12.95 0.665 13.105 0.665 9.9 15.7 755.0 117.0 5.45 250.00 38.10 3.13 131.00 58.50 4.12HP13.x 73. 21.6 12.75 0.565 13.005 0.565 11.5 18.4 630.0 98.8 5.40 207.00 31.90 3.10 110.00 48.80 2.54HP13.x 60. 17.5 12.54 0.460 12.900 0.460 14.0 22.7 503.0 80.3 5.36 165.00 25.50 3.07 89.00 39.00 1.39HP12.x 84. 24.6 12.28 0.685 12.295 0.685 9.0 14.2 650.0 106.0 5.14 213.00 34.60 2.94 120.00 53.20 4.24HP12.x 74. 21.8 12.13 0.605 12.215 0.610 10.0 16.0 569.0 93.8 5.11 186.00 30.40 2.92 105.00 46.60 2.98HP12.x 63. 18.4 11.94 0.515 12.125 0.515 11.8 18.9 472.0 79.1 5.06 153.00 25.30 2.88 88.30 38.70 1.83HP12.x 53. 15.5 11.78 0.435 12.045 0.435 13.8 22.3 393.0 66.8 5.03 127.00 21.10 2.86 74.00 32.20 1.12HP10.x 57. 16.8 9.99 0.565 10.225 0.565 9.0 13.9 294.0 58.8 4.18 101.00 19.70 2.45 66.50 30.30 1.97HP10.x 42. 12.4 9.70 0.415 10.075 0.420 12.0 18.9 210.0 43.4 4.13 71.70 14.20 2.41 48.30 21.80 0.81HP 8.x 36. 10.6 8.02 0.445 8.155 0.445 9.2 14.2 119.0 29.8 3.36 40.30 9.88 1.95 33.60 15.20 0.77



Designation A wgt Ix Sx rx Iy Sy ry Zx Zy J

in2 k/ft in4 in3 in in4 in3 in in3 in3 in4

L 8.0x8.0x1.125 16.70 56.90 98.0 17.5 2.42 98.00 17.50 2.42 31.60 31.60 7.13L 8.0x8.0x1.000 15.00 51.00 89.0 15.8 2.44 89.00 15.80 2.44 28.50 28.50 5.08L 8.0x8.0x0.875 13.20 45.00 79.6 14.0 2.45 79.60 14.00 2.45 25.30 25.30 3.46L 8.0x8.0x0.750 11.40 38.90 69.7 12.2 2.47 69.70 12.20 2.47 22.00 22.00 2.21L 8.0x8.0x0.625 9.61 32.70 59.4 10.3 2.49 59.40 10.30 2.49 18.60 18.60 1.30L 8.0x8.0x0.563 8.68 29.60 54.1 9.3 2.50 54.10 9.34 2.50 16.80 16.80 0.96L 8.0x8.0x0.500 7.75 26.40 48.6 8.4 2.50 48.60 8.36 2.50 15.10 15.10 0.68L 8.0x6.0x1.000 13.00 44.20 80.8 15.1 2.49 38.80 8.92 1.73 27.30 16.20 4.35L 8.0x6.0x0.875 11.50 39.10 72.3 13.4 2.51 34.90 7.94 1.74 24.20 14.40 2.96L 8.0x6.0x0.750 9.94 33.80 63.4 11.7 2.53 30.70 6.92 1.76 21.10 12.50 1.90L 8.0x6.0x0.625 8.36 28.50 54.1 9.9 2.54 26.30 5.88 1.77 17.90 10.50 1.12L 8.0x6.0x0.563 7.56 25.70 49.3 8.9 2.55 24.00 5.34 1.78 16.20 9.52 0.82L 8.0x6.0x0.500 6.75 23.00 44.3 8.0 2.56 21.70 4.79 1.79 14.50 8.51 0.58L 8.0x6.0x0.438 5.93 20.20 39.2 7.1 2.57 19.30 4.23 1.80 12.80 7.50 0.40L 8.0x4.0x1.000 11.00 37.40 69.6 14.1 2.52 11.60 3.94 1.03 24.30 7.72 3.68L 8.0x4.0x0.750 8.44 28.70 54.9 10.9 2.55 9.36 3.07 1.05 18.90 5.81 1.61L 8.0x4.0x0.563 6.43 21.90 42.8 8.4 2.58 7.43 2.38 1.07 14.50 4.38 0.70L 8.0x4.0x0.500 5.75 19.60 38.5 7.5 2.59 6.74 2.15 1.08 13.00 3.90 0.50L 7.0x4.0x0.750 7.69 26.20 37.8 8.4 2.22 9.05 3.03 1.09 14.80 5.65 1.47L 7.0x4.0x0.625 6.48 22.10 32.4 7.1 2.24 7.84 2.58 1.10 12.60 4.74 0.87L 7.0x4.0x0.500 5.25 17.90 26.7 5.8 2.25 6.53 2.12 1.11 10.30 3.83 0.46L 7.0x4.0x0.375 3.98 13.60 20.6 4.4 2.27 5.10 1.63 1.13 7.87 2.90 0.20L 6.0x6.0x1.000 11.00 37.40 35.5 8.6 1.80 35.50 8.57 1.80 15.50 15.50 3.68L 6.0x6.0x0.875 9.73 33.10 31.9 7.6 1.81 31.90 7.63 1.81 13.80 13.80 2.51L 6.0x6.0x0.750 8.44 28.70 28.2 6.7 1.83 28.20 6.66 1.83 12.00 12.00 1.61L 6.0x6.0x0.625 7.11 24.20 24.2 5.7 1.84 24.20 5.66 1.84 10.20 10.20 0.95L 6.0x6.0x0.563 6.43 21.90 22.1 5.1 1.85 22.10 5.14 1.85 9.26 9.26 0.70L 6.0x6.0x0.500 5.75 19.60 19.9 4.6 1.86 19.90 4.61 1.86 8.31 8.31 0.50L 6.0x6.0x0.438 5.06 17.20 17.7 4.1 1.87 17.70 4.08 1.87 7.34 7.34 0.34L 6.0x6.0x0.375 4.36 14.90 15.4 3.5 1.88 15.40 3.53 1.88 6.35 6.35 0.22L 6.0x6.0x0.313 3.65 12.40 13.0 3.0 1.89 13.00 2.97 1.89 5.35 5.35 0.13L 6.0x4.0x0.875 7.98 27.20 27.7 7.2 1.86 9.75 3.39 1.11 12.70 6.31 2.07L 6.0x4.0x0.750 6.94 23.60 24.5 6.3 1.88 8.68 2.97 1.12 11.20 5.47 1.33L 6.0x4.0x0.625 5.86 20.00 21.1 5.3 1.90 7.52 2.54 1.13 9.51 4.62 0.79L 6.0x4.0x0.563 5.31 18.10 19.3 4.8 1.90 6.91 2.31 1.14 8.66 4.19 0.58L 6.0x4.0x0.500 4.75 16.20 17.4 4.3 1.91 6.27 2.08 1.15 7.78 3.75 0.42L 6.0x4.0x0.438 4.18 14.30 15.5 3.8 1.92 5.60 1.85 1.16 6.88 3.30 0.28L 6.0x4.0x0.375 3.61 12.30 13.5 3.3 1.93 4.90 1.60 1.17 5.97 2.85 0.18L 6.0x4.0x0.313 3.03 10.30 11.4 2.8 1.94 4.18 1.35 1.17 5.03 2.40 0.11L 6.0x3.5x0.500 4.50 15.30 16.6 4.2 1.92 4.25 1.59 0.97 7.50 2.91 0.40L 6.0x3.5x0.375 3.42 11.70 12.9 3.2 1.94 3.34 1.23 0.99 5.76 2.20 0.17L 6.0x3.5x0.313 2.87 9.80 10.9 2.7 1.95 2.85 1.04 1.00 4.85 1.85 0.10L 5.0x5.0x0.875 7.98 27.20 17.8 5.2 1.49 17.80 5.17 1.49 9.33 9.33 2.07L 5.0x5.0x0.750 6.94 23.60 15.7 4.5 1.51 15.70 4.53 1.51 8.16 8.16 1.33L 5.0x5.0x0.625 5.86 20.00 13.6 3.9 1.52 13.60 3.86 1.52 6.95 6.95 0.79L 5.0x5.0x0.500 4.75 16.20 11.3 3.2 1.54 11.30 3.16 1.54 5.68 5.68 0.42L 5.0x5.0x0.438 4.18 14.30 10.0 2.8 1.55 10.00 2.79 1.55 5.03 5.03 0.28L 5.0x5.0x0.375 3.61 12.30 8.7 2.4 1.56 8.74 2.42 1.56 4.36 4.36 0.18L 5.0x5.0x0.313 3.03 10.30 7.4 2.0 1.57 7.42 2.04 1.57 3.68 3.68 0.11L 5.0x3.5x0.750 5.81 19.80 13.9 4.3 1.55 5.55 2.22 0.98 7.65 4.10 1.11L 5.0x3.5x0.625 4.92 16.80 12.0 3.7 1.56 4.83 1.90 0.99 6.55 3.47 0.66L 5.0x3.5x0.500 4.00 13.60 10.0 3.0 1.58 4.05 1.56 1.01 5.38 2.83 0.35L 5.0x3.5x0.438 3.53 12.00 8.9 2.6 1.59 3.63 1.39 1.01 4.77 2.49 0.24L 5.0x3.5x0.375 3.05 10.40 7.8 2.3 1.60 3.18 1.21 1.02 4.14 2.16 0.15L 5.0x3.5x0.313 2.56 8.70 6.6 1.9 1.61 2.72 1.02 1.03 3.49 1.82 0.09L 5.0x3.5x0.250 2.06 7.00 5.4 1.6 1.62 2.23 0.83 1.04 2.83 1.47 0.05L 5.0x3.0x0.625 4.61 15.70 11.4 3.5 1.57 3.06 1.39 0.81 6.27 2.61 0.61L 5.0x3.0x0.500 3.75 12.80 9.4 2.9 1.59 2.58 1.15 0.83 5.16 2.11 0.32L 5.0x3.0x0.438 3.31 11.30 8.4 2.6 1.60 2.32 1.02 0.84 4.57 1.86 0.22L 5.0x3.0x0.375 2.86 9.80 7.4 2.2 1.61 2.04 0.89 0.85 3.97 1.60 0.14L 5.0x3.0x0.313 2.40 8.20 6.3 1.9 1.61 1.75 0.75 0.85 3.36 1.35 0.08L 5.0x3.0x0.250 1.94 6.60 5.1 1.5 1.62 1.44 0.61 0.86 2.72 1.09 0.04L 4.0x4.0x0.750 5.44 18.50 7.7 2.8 1.19 7.67 2.81 1.19 5.07 5.07 1.02L 4.0x4.0x0.625 4.61 15.70 6.7 2.4 1.20 6.66 2.40 1.20 4.33 4.33 0.61L 4.0x4.0x0.500 3.75 12.80 5.6 2.0 1.22 5.56 1.97 1.22 3.56 3.56 0.32L 4.0x4.0x0.438 3.31 11.30 5.0 1.8 1.23 4.97 1.75 1.23 3.16 3.16 0.22L 4.0x4.0x0.375 2.86 9.80 4.4 1.5 1.23 4.36 1.52 1.23 2.74 2.74 0.14L 4.0x4.0x0.313 2.40 8.20 3.7 1.3 1.24 3.71 1.29 1.24 2.32 2.32 0.08L 4.0x4.0x0.250 1.94 6.60 3.0 1.0 1.25 3.04 1.05 1.25 1.88 1.88 0.04L 4.0x3.5x0.500 3.50 11.90 5.3 1.9 1.23 3.79 1.52 1.04 3.50 2.73 0.30L 4.0x3.5x0.438 3.09 10.60 4.8 1.7 1.24 3.40 1.35 1.05 3.11 2.42 0.21L 4.0x3.5x0.375 2.67 9.10 4.2 1.5 1.25 2.95 1.16 1.06 2.71 2.11 0.13L 4.0x3.5x0.313 2.25 7.70 3.6 1.3 1.26 2.55 0.99 1.07 2.29 1.78 0.08L 4.0x3.5x0.250 1.81 6.20 2.9 1.0 1.27 2.09 0.81 1.07 1.86 1.44 0.04L 4.0x3.0x0.500 3.25 11.10 5.1 1.9 1.25 2.42 1.12 0.86 3.41 2.03 0.28L 4.0x3.0x0.438 2.87 9.80 4.5 1.7 1.25 2.18 0.99 0.87 3.03 1.79 0.19L 4.0x3.0x0.375 2.48 8.50 4.0 1.5 1.26 1.92 0.87 0.88 2.64 1.56 0.12L 4.0x3.0x0.313 2.09 7.20 3.4 1.2 1.27 1.65 0.73 0.89 2.23 1.31 0.07L 4.0x3.0x0.250 1.69 5.80 2.8 1.0 1.28 1.36 0.60 0.90 1.82 1.06 0.04L 3.5x3.5x0.500 3.25 11.10 3.6 1.5 1.06 3.64 1.49 1.06 2.68 2.68 0.28L 3.5x3.5x0.438 2.87 9.80 3.3 1.3 1.07 3.26 1.32 1.07 2.38 2.38 0.19L 3.5x3.5x0.375 2.48 8.50 2.9 1.1 1.07 2.87 1.15 1.07 2.08 2.08 0.12L 3.5x3.5x0.313 2.09 7.20 2.5 1.0 1.08 2.45 0.98 1.08 1.76 1.76 0.07L 3.5x3.5x0.250 1.69 5.80 2.0 0.8 1.09 2.01 0.79 1.09 1.43 1.43 0.04L 3.5x3.0x0.500 3.00 10.20 3.5 1.5 1.07 2.33 1.10 0.88 2.63 1.98 0.26L 3.5x3.0x0.438 2.65 9.10 3.1 1.3 1.08 2.09 0.98 0.89 2.34 1.76 0.18L 3.5x3.0x0.375 2.30 7.90 2.7 1.1 1.09 1.85 0.85 0.90 2.04 1.53 0.11L 3.5x3.0x0.313 1.93 6.60 2.3 1.0 1.10 1.58 0.72 0.90 1.73 1.30 0.07L 3.5x3.0x0.250 1.56 5.40 1.9 0.8 1.11 1.30 0.59 0.91 1.41 1.05 0.04L 3.5x2.5x0.500 2.75 9.40 3.2 1.4 1.09 1.36 0.76 0.70 2.53 1.40 0.23L 3.5x2.5x0.438 2.43 8.30 2.9 1.3 1.09 1.23 0.68 0.71 2.26 1.24 0.16L 3.5x2.5x0.375 2.11 7.20 2.6 1.1 1.10 1.09 0.59 0.72 1.97 1.07 0.10L 3.5x2.5x0.313 1.78 6.10 2.2 0.9 1.11 0.94 0.50 0.73 1.67 0.91 0.06L 3.5x2.5x0.250 1.44 4.90 1.8 0.8 1.12 0.78 0.41 0.74 1.36 0.74 0.03L 3.0x3.0x0.500 2.75 9.40 2.2 1.1 0.90 2.22 1.07 0.90 1.93 1.93 0.23L 3.0x3.0x0.438 2.43 8.30 2.0 1.0 0.90 1.99 0.95 0.90 1.72 1.72 0.16L 3.0x3.0x0.375 2.11 7.20 1.8 0.8 0.91 1.76 0.83 0.91 1.50 1.50 0.10L 3.0x3.0x0.313 1.78 6.10 1.5 0.7 0.92 1.51 0.71 0.92 1.27 1.27 0.06L 3.0x3.0x0.250 1.44 4.90 1.2 0.6 0.93 1.24 0.58 0.93 1.04 1.04 0.03L 3.0x3.0x0.188 1.09 3.71 1.0 0.4 0.94 0.96 0.44 0.94 0.79 0.79 0.01L 3.0x2.5x0.500 2.50 8.50 2.1 1.0 0.91 1.30 0.74 0.72 1.88 1.35 0.21L 3.0x2.5x0.438 2.21 7.60 1.9 0.9 0.92 1.18 0.66 0.73 1.68 1.20 0.15L 3.0x2.5x0.375 1.92 6.60 1.7 0.8 0.93 1.04 0.58 0.74 1.47 1.05 0.09L 3.0x2.5x0.313 1.62 5.60 1.4 0.7 0.94 0.90 0.49 0.74 1.25 0.89 0.06L 3.0x2.5x0.250 1.31 4.50 1.2 0.6 0.94 0.74 0.40 0.75 1.02 0.72 0.03L 3.0x2.5x0.188 1.00 3.39 0.9 0.4 0.95 0.58 0.31 0.76 0.78 0.55 0.01L 3.0x2.0x0.500 2.25 7.70 1.9 1.0 0.92 0.67 0.47 0.55 1.78 0.89 0.19L 3.0x2.0x0.438 2.00 6.80 1.7 0.9 0.93 0.61 0.42 0.55 1.59 0.79 0.13L 3.0x2.0x0.375 1.73 5.90 1.5 0.8 0.94 0.54 0.37 0.56 1.40 0.68 0.09L 3.0x2.0x0.313 1.46 5.00 1.3 0.7 0.95 0.47 0.32 0.57 1.19 0.58 0.05L 3.0x2.0x0.250 1.19 4.10 1.1 0.5 0.96 0.39 0.26 0.57 0.97 0.47 0.03L 3.0x2.0x0.188 0.90 3.07 0.8 0.4 0.97 0.31 0.20 0.58 0.75 0.36 0.01L 2.5x2.5x0.500 2.25 7.70 1.2 0.7 0.74 1.23 0.72 0.74 1.31 1.31 0.19L 2.5x2.5x0.375 1.73 5.90 1.0 0.6 0.75 0.98 0.57 0.75 1.02 1.02 0.08L 2.5x2.5x0.313 1.46 5.00 0.8 0.5 0.76 0.85 0.48 0.76 0.87 0.87 0.05L 2.5x2.5x0.250 1.19 4.10 0.7 0.4 0.77 0.70 0.39 0.77 0.71 0.71 0.03L 2.5x2.5x0.188 0.90 3.07 0.5 0.3 0.78 0.55 0.30 0.78 0.55 0.55 0.01L 2.5x2.0x0.375 1.55 5.30 0.9 0.5 0.77 0.51 0.36 0.58 0.99 0.66 0.07L 2.5x2.0x0.313 1.31 4.50 0.8 0.5 0.78 0.45 0.31 0.58 0.84 0.56 0.04



3.6.1 ASCII File with Steel Section Properties

Available is an ASCII file (PC/DOS or Sun/UNIX (/usr/src/public/aisc) containing the followingsection properties (All dimensions are inches and kips).

A AreaBF Width of flangeBTF ”bf/2tf” the ratio of the flange width to twice the flange thicknessCW Cw, warping constantD Depth, actualDAF ”d/Af” the ratio of the depth to the compression flange areaDI Inner diameter of pipeDN Nominal depthDSG Shape designation i.e. ”W” or ”C”DTW ”d/tw” the ratio of the depth to the web thicknessEO ”eo” distance from outside face of web to shear centerFYP ”Fy’” the theoretical maximum yield stress based on the width-thickness ratio of 1/2 the

unstiffened compression flange, beyond which a particular shape is not compact. (Specifiedas 0 if greater than 65 ksi)

FYPPPA ”Fy”’” (ASD) the theoretical maximum yield stress based on the depth-thickness ratio ofthe web below which a particular shape may be considered compact for any condition ofcombined bending and axial stresses. (Specified as 0 if greater than 65 ksi)

FYPPPL ”Fy”’” (LRFD) see aboveH ”H” flexural constant LRFD Formula (A-E3-9)H38 ”H” flexural constant, double angles 3/8” back to backH34 ”H” flexural constant, double angles 3/4” back to backHL Horizontal leg or side dimensionHTW ”hc/tw”, ratio of assumed web depth for stability to thickess of webJSHP JUMBO SHAPE (”J”) see note belowNT Tensile group number per ASTM A6NX1 ”X1” beam buckling factor LRFD Formula (F1-8)OD Outer diameter of pipeQF ”Qf” statical moment at point in flangeQW ”Qw” statical moment at mid-depth



RA Minimum fillet radius, detailing value (in.); flange toe value for M, S, C, MC shapesRI Minimum fillet radius, designRO ”ro” shear center coordinateRO38 ”ro” double angles 3/8” back to backRO34 ”ro” double angles 3/4” back to backRT The radius of gyration of a section comprising the compression flange plus 1/3 of the com-

pression web area, taken about an axis in the plane of the webRX ”rx” radius of gyration about axis XXRY ”ry” radius of gyration about axis YYRY38 ”ry” double angles 3/8” back to backRY34 ”ry” double angles 3/4” back to backRZ ”rz” radius of gyration about principal axisSX ”Sx” elastic section modulus about axis XXSY ”Sy” elastic section modulus about axis YYSW ”Sw” warping statical momentT Thickness of leg or wallTF Thickness of flangeTW Thickness of webVL Vertical leg or side dimensionWGT Nominal weightWNO ”Wno” normalized warping functionXB ”x” centroid of section in X direction from outer webXI ”Ix” moment of inertia about axis XXXJ ”J” torsional constantXK ”k” distance from outer face of flange to web toe fillet of rolled shapeX2 ”X2” beam buckling factor LRFD Formula (F1-9)YB ”y” centroid of section in Y direction from outer top flange faceYI ”Iy” moment of inertia about axis YYZX ”Zx” plastic modulus about axis XXZY ”Zy” plastic modulus about axis YY

program form

c

c test program to open the AISC1.PRN file, read a single record

c and print the data from that record to the screen.

c

c aisc1.bin contains contains 658 records (sections) with

c 38 variables in each record.

c

character*2 dsg, jshp

real dn, wgt, a, d, tw, bf, tf, xk,

1 btf, fyp, htw, dtw, fypppl, fypppa, x1, x2, rt, daf,

2 ri, ra, nt, xi, sx, rx, yi, sy, ry, zx, zy, xj, cw,

3 wno, sw, qf, qw, ro

c ---------------------------------

c open aisc file

c ---------------------------------

open(unit=1,file=’aisc1.bin’,

1 status=’old’,access=’direct’,recl=152)

c -----------------------------------

c prompt user to select record (line)

c -----------------------------------

100 write(*,*) ’ Enter line number to read: (0 to quit) >’

read(*,*) line

if(line .eq. 0) stop

if(line.gt.658)then

write(*,*)’ Invalid line number, must be less than 658’

goto 100

endif

c ------------------------------------

c read a single record from aisc file

c ------------------------------------


Draft3.7 Joists 3–19

read(1,rec=line) dsg, dn, wgt, jshp, a, d, tw, bf, tf, xk,




c -------------------------

c write data to screen

c -------------------------

write(*,999,err=1000) dsg, dn, wgt, jshp, a, d, tw, bf, tf, xk,




c ------

c goto beginning

c ------

goto 100

c---------------------------------------------------------

c error message if read attempt is past last record (658)

c---------------------------------------------------------

1000 write(*,*) ’ line number is past last record (658) ’

goto 100

999 format(a2,2x,f3.0,2x,f4.0,a2,6(2x,1pe10.3))

end

3.7 Joists

41 Table 3.3 list the Joist properties.



4" 4"4"

Spa

n

[De

sign

Le

ng

th =

Sp

an

– 0

.33

FT.]

Figure 3.9: prefabricated Steel Joists


Draft3.7 Joists 3–21

Joint 8K1 10K1 12K1 12K3 12K5 14K1 14K3 14K4 14K6 16K2 16K3 16K4 16K5 16K6 16K7 16K9Desig.Depth(in.)

8 10 12 12 12 14 14 14 14 16 16 16 16 16 16 16

≈ W 5.1 5 5 5.7 7.1 5.2 6 6.7 7.7 5.5 6.3 7 7.5 8.1 8.6 10.0(lbs/ft)Span(ft.)8 550

5509 550

55010 550 550

480 55011 532 550

377 54212 444 550 550 550 550

288 455 550 550 55013 377 479 550 550 550

225 363 510 510 51014 324 412 500 550 550 550 550 550 550

179 289 425 463 463 550 550 550 55015 281 358 434 543 550 511 550 550 550

145 234 344 428 434 475 507 507 50716 246 313 380 476 550 448 550 550 550 550 550 550 550 550 550 550

119 192 282 351 396 390 467 467 467 550 550 550 550 550 550 55017 277 336 420 550 395 495 550 550 512 550 550 550 550 550 550

159 234 291 366 324 404 443 443 488 526 526 526 526 526 52618 246 299 374 507 352 441 530 550 456 508 550 550 550 550 550

134 197 245 317 272 339 397 408 409 456 490 490 490 490 49019 221 268 335 454 315 395 475 550 408 455 547 550 550 550 550

113 167 207 269 230 287 336 383 347 386 452 455 455 455 45520 199 241 302 409 284 356 428 525 368 410 493 550 550 550 550

97 142 177 230 197 246 287 347 297 330 386 426 426 426 42621 218 273 370 257 322 388 475 333 371 447 503 548 550 550

123 153 198 170 212 248 299 255 285 333 373 405 406 40622 199 249 337 234 293 353 432 303 337 406 458 498 550 550

106 132 172 147 184 215 259 222 247 289 323 351 385 38523 181 227 308 214 268 322 395 277 308 371 418 455 507 550

93 116 150 128 160 188 226 194 216 252 282 307 339 36324 166 208 282 196 245 295 362 254 283 340 384 418 465 550

81 101 132 113 141 165 199 170 189 221 248 269 298 34625 180 226 272 334 234 260 313 353 384 428 514

100 124 145 175 150 167 195 219 238 263 31126 166 209 251 308 216 240 289 326 355 395 474

88 110 129 56 133 148 173 194 211 233 27627 154 193 233 285 200 223 268 302 329 366 439

79 98 115 139 119 132 155 173 188 208 24628 143 180 216 265 186 207 249 281 306 340 408

70 88 103 124 106 118 138 155 168 186 22029 173 193 232 261 285 317 380

95 106 124 139 151 167 19830 161 180 216 244 266 296 355

86 96 112 126 137 151 17831 151 168 203 228 249 277 332

78 87 101 114 124 137 16132 142 158 190 214 233 259 311

71 79 92 103 112 124 147

Table 3.3: Joist Properties


Draft

Chapter 4

EQUILIBRIUM & REACTIONS

To every action there is an equal and opposite reaction.

Newton’s third law of motion

4.1 Introduction

1 In the analysis of structures (hand calculations), it is often easier (but not always necessary) to startby determining the reactions.

2 Once the reactions are determined, internal forces are determined next; finally, deformations (deflec-tions and rotations) are determined last1.

3 Reactions are necessary to determine foundation load.

4 Depending on the type of structures, there can be different types of support conditions, Fig. 4.1.

Figure 4.1: Types of Supports

1This is the sequence of operations in the flexibility method which lends itself to hand calculation. In the stiffnessmethod, we determine displacements firsts, then internal forces and reactions. This method is most suitable to computerimplementation.

Draft4–2 EQUILIBRIUM & REACTIONS

Roller: provides a restraint in only one direction in a 2D structure, in 3D structures a roller may providerestraint in one or two directions. A roller will allow rotation.

Hinge: allows rotation but no displacements.

Fixed Support: will prevent rotation and displacements in all directions.

4.2 Equilibrium

5 Reactions are determined from the appropriate equations of static equilibrium.

6 Summation of forces and moments, in a static system must be equal to zero2.

7 In a 3D cartesian coordinate system there are a total of 6 independent equations of equilibrium:ΣFx = ΣFy = ΣFz = 0ΣMx = ΣMy = ΣMz = 0 (4.1)

8 In a 2D cartesian coordinate system there are a total of 3 independent equations of equilibrium:

ΣFx = ΣFy = ΣMz = 0 (4.2)

9 For reaction calculations, the externally applied load may be reduced to an equivalent force3.

10 Summation of the moments can be taken with respect to any arbitrary point.

11 Whereas forces are represented by a vector, moments are also vectorial quantities and are representedby a curved arrow or a double arrow vector.

12 Not all equations are applicable to all structures, Table 4.1

Structure Type EquationsBeam, no axial forces ΣFy ΣMz

2D Truss, Frame, Beam ΣFx ΣFy ΣMz

Grid ΣFz ΣMx ΣMy

3D Truss, Frame ΣFx ΣFy ΣFz ΣMx ΣMy ΣMz

Alternate SetBeams, no axial Force ΣMA

z ΣMBz

2 D Truss, Frame, Beam ΣFx ΣMAz ΣMB

z

ΣMAz ΣMB

z ΣMCz

Table 4.1: Equations of Equilibrium

13 The three conventional equations of equilibrium in 2D: ΣFx,ΣFy and ΣMz can be replaced by theindependent moment equations ΣMA

z , ΣMBz , ΣMC

z provided that A, B, and C are not colinear.

14 It is always preferable to check calculations by another equation of equilibrium.

15 Before you write an equation of equilibrium,

1. Arbitrarily decide which is the +ve direction

2. Assume a direction for the unknown quantities

3. The right hand side of the equation should be zero2In a dynamic system ΣF = ma where m is the mass and a is the acceleration.3However for internal forces (shear and moment) we must use the actual load distribution.


Draft4.3 Equations of Conditions 4–3

If your reaction is negative, then it will be in a direction opposite from the one assumed.

16 Summation of all external forces (including reactions) is not necessarily zero (except at hinges and atpoints outside the structure).

17 Summation of external forces is equal and opposite to the internal ones. Thus the net force/momentis equal to zero.

18 The external forces give rise to the (non-zero) shear and moment diagram.

4.3 Equations of Conditions

19 If a structure has an internal hinge (which may connect two or more substructures), then this willprovide an additional equation (ΣM = 0 at the hinge) which can be exploited to determine the reactions.

20 Those equations are often exploited in trusses (where each connection is a hinge) to determine reac-tions.

21 In an inclined roller support with Sx and Sy horizontal and vertical projection, then the reactionR would have, Fig. 4.2.

RxRy

=SySx

(4.3)

Figure 4.2: Inclined Roller Support

4.4 Static Determinacy

22 In statically determinate structures, reactions depend only on the geometry, boundary conditions andloads.

23 If the reactions can not be determined simply from the equations of static equilibrium (and equationsof conditions if present), then the reactions of the structure are said to be statically indeterminate.

24 the degree of static indeterminacy is equal to the difference between the number of reactions andthe number of equations of equilibrium, Fig. 4.3.

25 Failure of one support in a statically determinate system results in the collapse of the structures.Thus a statically indeterminate structure is safer than a statically determinate one.

26 For statically indeterminate structures4, reactions depend also on the material properties (e.g. Young’sand/or shear modulus) and element cross sections (e.g. length, area, moment of inertia).

4Which will be studied in CVEN3535.



Figure 4.3: Examples of Static Determinate and Indeterminate Structures

Example 4-1: Statically Indeterminate Cable Structure

A rigid plate is supported by two aluminum cables and a steel one. Determine the force in eachcable5.

If the rigid plate supports a load P, determine the stress in each of the three cables. Solution:

1. We have three unknowns and only two independent equations of equilibrium. Hence the problemis statically indeterminate to the first degree.

ΣMz = 0; ⇒ P leftAl = P right

Al

ΣFy = 0; ⇒ 2PAl + PSt = P

Thus we effectively have two unknowns and one equation.

2. We need to have a third equation to solve for the three unknowns. This will be derived from thecompatibility of the displacements in all three cables, i.e. all three displacements must beequal:

σ = PA

ε = ∆LL

ε = σE

⇒ ∆L =

PL

AE5This example problem will be the only statically indeterminate problem analyzed in CVEN3525.


Draft4.5 Geometric Instability 4–5

PAlL

EAlAAl︸︷︷︸∆Al

=PStL

EStASt︸︷︷︸∆St

⇒ PAlPSt

=(EA)Al(EA)St

or −(EA)StPAl + (EA)AlPSt = 0

3. Solution of this system of two equations with two unknowns yield:[2 1

−(EA)St (EA)Al

]{PAlPSt

}={P0

}⇒{PAlPSt

}=[

2 1−(EA)St (EA)Al

]−1{P0

}=

12(EA)Al + (EA)St︸︷︷︸

Determinant

[(EA)Al −1(EA)St 2

]{P0

}

4.5 Geometric Instability

27 The stability of a structure is determined not only by the number of reactions but also by theirarrangement.

28 Geometric instability will occur if:

1. All reactions are parallel and a non-parallel load is applied to the structure.

2. All reactions are concurrent, Fig. ??.

Figure 4.4: Geometric Instability Caused by Concurrent Reactions

3. The number of reactions is smaller than the number of equations of equilibrium, that is a mech-anism is present in the structure.

29 Mathematically, this can be shown if the determinant of the equations of equilibrium is equal tozero (or the equations are inter-dependent).

4.6 Examples

30 Examples of reaction calculation will be shown next. Each example has been carefully selected as itbrings a different “twist” from the preceding one. Some of those same problems will be revisited laterfor the determination of the internal forces and/or deflections. Many of those problems are taken fromProf. Gerstle textbok Basic Structural Analysis.

Example 4-2: Simply Supported Beam



Determine the reactions of the simply supported beam shown below.

Solution:The beam has 3 reactions, we have 3 equations of static equilibrium, hence it is statically determinate.

(+✲ )ΣFx = 0; ⇒ Rax − 36 k = 0(+ ✻)ΣFy = 0; ⇒ Ray +Rdy − 60 k− (4) k/ft(12) ft = 0

(+ �✁✛)ΣM c

z = 0; ⇒ 12Ray − 6Rdy − (60)(6) = 0

or 1 0 0

0 1 10 12 −6

RaxRayRdy

=

36108360

⇒

RaxRayRdy

=

36 k

56 k

52 k

Alternatively we could have used another set of equations:

(+ �✁✛)ΣMa

z = 0; (60)(6) + (48)(12)− (Rdy)(18) = 0 ⇒ Rdy = 52 k ✻(+ �

✁✛)ΣMdz = 0; (Ray)(18)− (60)(12)− (48)(6) = 0 ⇒ Ray = 56 k ✻

Check:(+ ✻)ΣFy = 0; ; 56− 52− 60− 48 = 0

√

Example 4-3: Parabolic Load

Determine the reactions of a simply supported beam of length L subjected to a parabolic loadw = w0

(xL

)2


Draft4.6 Examples 4–7

Solution:Since there are no axial forces, there are two unknowns and two equations of equilibrium. Consideringan infinitesimal element of length dx, weight dW , and moment dM :

(+ �✁✛)ΣMa = 0;

∫ x=L

x=0

w0

( xL

)2︸︷︷︸

w

dx

︸︷︷︸dW

×x

︸︷︷︸dM︸︷︷︸

M

−(Rb)(L) = 0

⇒ Rb = 1Lw0

(L4

4L2

)= 1

4w0L

(+ ✻)ΣFy = 0; Ra +14w0L︸︷︷︸Rb

−∫ x=L

x=0

w0

( xL

)2dx = 0

⇒ Ra = w0L2

L3

3 − 14w0L = 1

12w0L

Example 4-4: Three Span Beam

Determine the reactions of the following three spans beam

Solution:We have 4 unknowns (Rax, Ray, Rcy and Rdy), three equations of equilibrium and one equation ofcondition (ΣMb = 0), thus the structure is statically determinate.



1. Isolating ab:ΣM �

✁✛b = 0; (9)(Ray)− (40)(5) = 0 ⇒ Ray = 22.2 k ✻(+ �

✁✛)ΣMa = 0; (40)(4)− (S)(9) = 0 ⇒ S = 17.7 k ✻ΣFx = 0; ⇒ Rax = 30 k ✛

2. Isolating bd:

(+ �✁✛)ΣMd = 0; −(17.7)(18)− (40)(15)− (4)(8)(8)− (30)(2) +Rcy(12) = 0

⇒ Rcy = 1,23612 = 103 k ✻

(+ �✁✛)ΣMc = 0; −(17.7)(6)− (40)(3) + (4)(8)(4) + (30)(10)−Rdy(12) = 0

⇒ Rdy = 201.312 = 16.7 k ✻

3. CheckΣFy = 0; ✻; 22.2− 40− 40 + 103− 32− 30 + 16.7 = 0

√

Example 4-5: Three Hinged Gable Frame

The three-hinged gable frames spaced at 30 ft. on center. Determine the reactions components on theframe due to: 1) Roof dead load, of 20 psf of roof area; 2) Snow load, of 30 psf of horizontal projection;3) Wind load of 15 psf of vertical projection. Determine the critical design values for the vertical andhorizontal reactions.



Solution:

1. Due to symmetry, we will consider only the dead load on one side of the frame.

2. Due to symmetry, there is no vertical force transmitted by the hinge for snow and dead load.

3. Roof dead load per frame is

DL = (20) psf(30) ft(√

302 + 152)

ft1

1, 000lbs/k = 20.2 k ❄

4. Snow load per frame is

SL = (30) psf(30) ft(30) ft1

1, 000lbs/k = 27. k ❄

5. Wind load per frame (ignoring the suction) is

WL = (15) psf(30) ft(35) ft1

1, 000lbs/k = 15.75 k✲

6. There are 4 reactions, 3 equations of equilibrium and one equation of condition ⇒ staticallydeterminate.

7. The horizontal reaction H due to a vertical load V at midspan of the roof, is obtained by takingmoment with respect to the hinge

(+ �✁✛) ΣMC = 0; 15(V )− 30(V ) + 35(H) = 0 ⇒ H = 15V

35 = .429V

Substituting for roof dead and snow load we obtain

V ADL = V BDL = 20.2 k ✻HADL = HB

DL = (.429)(20.2) = 8.66 k✲V ASL = V BSL = 27. k ✻HASL = HB

SL = (.429)(27.) = 11.58 k✲

8. The reactions due to wind load are

(+ �✁✛)ΣMA = 0; (15.75)(20+152 )− V BWL(60) = 0 ⇒ V BWL = 4.60 k ✻

(+ �✁✛)ΣMC = 0; HB

WL(35)− (4.6)(30) = 0 ⇒ HBWL = 3.95 k ✛

(+✲ )ΣFx = 0; 15.75− 3.95−HAWL = 0 ⇒ HA

WL = 11.80 k ✛(+ ✻)ΣFy = 0; V BWL − V AWL = 0 ⇒ V AWL = −4.60 k ❄



9. Thus supports should be designed for

H = 8.66 k + 11.58 k + 3.95 k = 24.19 k

V = 20.7 k + 27.0 k + 4.60 k = 52.3 k

Example 4-6: Inclined Supports

Determine the reactions of the following two spans beam resting on inclined supports.

Solution:A priori we would identify 5 reactions, however we do have 2 equations of conditions (one at each inclinedsupport), thus with three equations of equilibrium, we have a statically determinate system.

(+ �✁✛)ΣMb = 0; (Ray)(20)− (40)(12)− (30)(6) + (44.72)(6)− (Rcy)(12) = 0

⇒ 20Ray = 12Rcy + 391.68(+✲ )ΣFx = 0; 3

4Ray − 22.36− 43Rcy = 0

⇒ Rcy = 0.5625Ray − 16.77

Solving for those two equations:[20 −12

0.5625 −1

]{RayRcy

}={

391.6816.77

}⇒{RayRcy

}={

14.37 k ✻−8.69 k ❄

}The horizontal components of the reactions at a and c are

Rax = 3414.37 = 10.78 k✲

Rcx = 438.69 = −11.59 k✲

Finally we solve for Rby

(+ �✁✛)ΣMa = 0; (40)(8) + (30)(14)− (Rby)(20) + (44.72)(26) + (8.69)(32) = 0

⇒ Rby = 109.04 k ✻

We check our results

(+ ✻)ΣFy = 0; 14.37− 40− 30 + 109.04− 44.72− 8.69 = 0√

(+✲ ) ΣFx = 0; 10.78− 22.36 + 11.59 = 0√


Draft4.7 Arches 4–11

4.7 Arches

31 See Section ??.


Draft

Chapter 5

TRUSSES

5.1 Introduction

5.1.1 Assumptions

1 Cables and trusses are 2D or 3D structures composed of an assemblage of simple one dimensionalcomponents which transfer only axial forces along their axis.

2 Cables can carry only tensile forces, trusses can carry tensile and compressive forces.

3 Cables tend to be flexible, and hence, they tend to oscillate and therefore must be stiffened.

4 Trusses are extensively used for bridges, long span roofs, electric tower, space structures.

5 For trusses, it is assumed that

1. Bars are pin-connected

2. Joints are frictionless hinges1.

3. Loads are applied at the joints only.

6 A truss would typically be composed of triangular elements with the bars on the upper chord undercompression and those along the lower chord under tension. Depending on the orientation of thediagonals, they can be under either tension or compression.

7 Fig. 5.1 illustrates some of the most common types of trusses.

8 It can be easily determined that in a Pratt truss, the diagonal members are under tension, while ina Howe truss, they are in compression. Thus, the Pratt design is an excellent choice for steel whosemembers are slender and long diagonal member being in tension are not prone to buckling. The verticalmembers are less likely to buckle because they are shorter. On the other hand the Howe truss is oftenpreferred for for heavy timber trusses.

9 In a truss analysis or design, we seek to determine the internal force along each member, Fig. 5.2

5.1.2 Basic Relations

Sign Convention: Tension positive, compression negative. On a truss the axial forces are indicated asforces acting on the joints.

Stress-Force: σ = PA

Stress-Strain: σ = Eε1In practice the bars are riveted, bolted, or welded directly to each other or to gusset plates, thus the bars are not free

to rotate and so-called secondary bending moments are developed at the bars. Another source of secondary momentsis the dead weight of the element.

Draft5–2 TRUSSES

Figure 5.1: Types of Trusses

Figure 5.2: Bridge Truss


Draft5.2 Trusses 5–3

Force-Displacement: ε = ∆LL

Equilibrium: ΣF = 0

5.2 Trusses

5.2.1 Determinacy and Stability

10 Trusses are statically determinate when all the bar forces can be determined from the equationsof statics alone. Otherwise the truss is statically indeterminate.

11 A truss may be statically/externally determinate or indeterminate with respect to the reactions (morethan 3 or 6 reactions in 2D or 3D problems respectively).

12 A truss may be internally determinate or indeterminate, Table 5.1.

13 If we refer to j as the number of joints, R the number of reactions and m the number of members,then we would have a total of m + R unknowns and 2j (or 3j) equations of statics (2D or 3D at eachjoint). If we do not have enough equations of statics then the problem is indeterminate, if we have toomany equations then the truss is unstable, Table 5.1.

2D 3DStatic Indeterminacy

External R > 3 R > 6Internal m+R > 2j m+ R > 3jUnstable m+R < 2j m+ R < 3j

Table 5.1: Static Determinacy and Stability of Trusses

14 If m < 2j − 3 (in 2D) the truss is not internally stable, and it will not remain a rigid body when it isdetached from its supports. However, when attached to the supports, the truss will be rigid.

15 Since each joint is pin-connected, we can apply ΣM = 0 at each one of them. Furthermore, summationof forces applied on a joint must be equal to zero.

16 For 2D trusses the external equations of equilibrium which can be used to determine the reactionsare ΣFX = 0, ΣFY = 0 and ΣMZ = 0. For 3D trusses the available equations are ΣFX = 0, ΣFY = 0,ΣFZ = 0 and ΣMX = 0, ΣMY = 0, ΣMZ = 0.

17 For a 2D truss we have 2 equations of equilibrium ΣFX = 0 and ΣFY = 0 which can be applied ateach joint. For 3D trusses we would have three equations: ΣFX = 0, ΣFY = 0 and ΣFZ = 0.

18 Fig. 5.3 shows a truss with 4 reactions, thus it is externally indeterminate. This truss has 6 joints(j = 6), 4 reactions (R = 4) and 9 members (m = 9). Thus we have a total of m + R = 9 + 4 = 13unknowns and 2× j = 2× 6 = 12 equations of equilibrium, thus the truss is statically indeterminate.

19 There are two methods of analysis for statically determinate trusses

1. The Method of joints

2. The Method of sections

5.2.2 Method of Joints

20 The method of joints can be summarized as follows

1. Determine if the structure is statically determinate

2. Compute all reactions


Draft5–4 TRUSSES

Figure 5.3: A Statically Indeterminate Truss

3. Sketch a free body diagram showing all joint loads (including reactions)

4. For each joint, and starting with the loaded ones, apply the appropriate equations of equilibrium(ΣFx and ΣFy in 2D; ΣFx, ΣFy and ΣFz in 3D).

5. Because truss elements can only carry axial forces, the resultant force (7F = 7Fx + 7Fy) must bealong the member, Fig. 5.4.

F

L=FxLx

=FyLy

(5.1)

21 Always keep track of the x and y components of a member force (Fx, Fy), as those might be neededlater on when considering the force equilibrium at another joint to which the member is connected.

Figure 5.4: X and Y Components of Truss Forces

22 This method should be used when all member forces must be determined.

23 In truss analysis, there is no sign convention. A member is assumed to be under tension (orcompression). If after analysis, the force is found to be negative, then this would imply that the wrongassumption was made, and that the member should have been under compression (or tension).

24 On a free body diagram, the internal forces are represented by arrow acting on the joints andnot as end forces on the element itself. That is for tension, the arrow is pointing away from the joint,and for compression toward the joint, Fig. 5.5.



Figure 5.5: Sign Convention for Truss Element Forces

Example 5-1: Truss, Method of Joints

Using the method of joints, analyze the following truss

Solution:

1. R = 3, m = 13, 2j = 16, and m+R = 2j√

2. We compute the reactions

(+ �✁✛)ΣME = 0; ⇒ (20 + 12)(3)(24) + (40 + 8)(2)(24) + (40)(24)−RAy(4)(24) = 0

⇒ RAy = 58 k ✻(+ ❄)ΣFy = 0; ⇒ 20 + 12 + 40 + 8 + 40− 58−REy = 0

⇒ REy = 62 k ✻


Draft5–6 TRUSSES

3. Consider each joint separately:

Node A: Clearly AH is under compression, and AB under tension.

(+ ✻)ΣFy = 0; ⇒ −FAHy + 58 = 0FAH = l

ly(FAHy )

ly = 32 l =√

322 + 242 = 40⇒ FAH = 40

32 (58) = 72.5 k Compression(+✲ )ΣFx = 0; ⇒ −FAHx + FAB = 0

FAB = LxLy

(FAHy ) = 2432 (58) = 43.5 k Tension

Node B:

(+✲ ) ΣFx = 0; ⇒ FBC = 43.5 k Tension(+ ✻)ΣFy = 0; ⇒ FBH = 20 k Tension

Node H:

(+✲ )ΣFx = 0; ⇒ FAHx − FHCx − FHGx = 043.5− 24√

242+322(FHC)− 24√

242+102(FHG) = 0

(+ ✻)ΣFy = 0; ⇒ FAHy + FHCy − 12− FHGy − 20 = 058 + 32√

242+322(FHC)− 12− 10√

242+102(FHG)− 20 = 0

This can be most conveniently written as[0.6 0.921−0.8 0.385

]{FHCFHG

}={ −7.5

52

}(5.2)

Solving we obtainFHC = −7.5 k TensionFHG = 52 k Compression



Node E:

ΣFy = 0; ⇒ FEFy = 62 ⇒ FEF =√242+322

32 (62) = 77.5 k

ΣFx = 0; ⇒ FED = FEFx ⇒ FED = 2432 (FEFy ) = 24

32 (62) = 46.5 k

The results of this analysis are summarized below

4. We could check our calculations by verifying equilibrium of forces at a node not previously used,such as D

5.2.2.1 Matrix Method

25 This is essentially the method of joints cast in matrix form2.

26 We seek to determine the Statics Matrix [B] such that[BFF BFRBRF BRR

]{FR

}={P0

}(5.3)

27 This method can be summarized as follows

1. Select a coordinate system

2. Number the joints and the elements separately2Writing a computer program for this method, would be an acceptable project.


Draft5–8 TRUSSES

3. Assume

(a) All member forces to be positive (i.e. tension)(b) All reactions to be positive

4. Compute the direction cosines at each node j and for each element e, Fig. 5.6

αβ

αβ + ve α

β+ ve+ ve

αβ

+ ve- ve

- ve- ve

- ve

X

Y

Figure 5.6: Direction Cosines

αej = LxL

βej = LyL

5. Write the two equations of equilibrium at each joint j in terms of the unknowns (member forcesand reactions), Fig. 5.7

e=F F eα

Fe

= LL

LL

= x

y

F ex

Fe

ye=F F eβ

Figure 5.7: Forces Acting on Truss Joint

(+✲ )ΣFx = 0; ⇒ Σ#ofelementse=1 αejFe +Rxj + Pxj = 0

(+ ✻)ΣFy = 0; ⇒ Σ#ofelementse=1 βejFe + Ryj + Pyj = 0

6. Invert the matrix to compute {F} and {R}

28 The advantage of this method, is that once the [B] matrix has been inverted, we can readily reanalyzethe same structure for different load cases. With the new design codes in which dead loads and liveloads are separately factored (Chapter 7), this method can save substantial reanalysis effort. Further-more, when deflections are determined by the virtual force method (Chapter 13), two analysis with twodifferent loads are required.

29 This method may be the only one appropriate to analyze statically determinate trusses which solu-tion’s defy the previous two methods, Fig. 5.8.

30 Element e connecting joint i to j will have αei = −αej , and βei = −βej31 The matrix [B] will have 2j rows and m+ r columns. It can only be inverted if ti is symmetric (i.e2j = m+ r, statically determinate).

32 An algorithm to implement this method in simple computer programs:



Figure 5.8: Complex Statically Determinate Truss

1. Prepare input data:

(a) Nodal information:

Node x(1) x(2) P (1) P (2) R(1) R(2)1 0. 0. 0. 0. 1 12 10. 0. 0. 0. 0 13 5. 5. 0. -10. 0 0

where x(1), x(2), P (1), and P (2) are the x and y coordinates; the x and y component ofapplied nodal load. R(1), R(2) correspond to the x and y boundary conditions, they will beset to 1 if there is a corresponding reaction, and 0 otherwise.

(b) Element Connectivity

Element Node(1) Node(2)1 1 22 2 33 3 1

2. Determine the size of the matrix (2 times the number of joints) and initialize a square matrix ofthis size to zero.

3. Assemble the first submnatrix of the Statics matrix

(a) Loop over each element (e), determine Lix, Liy (as measured from the first node), L, αei = Lix

L

βei = LiyL , αej = −αei , βej = −βei , where i and j are the end nodes of element e.

(b) Store αei in row 2i− 1 column e, βei in row 2i column e, αej in row 2j − 1 column e, βej in row2j column e.

4. Loop over each node:

(a) Store Rix (reaction boundary condition for node i along x) in row 2i − 1 column e + 2i− 1.Similarly, store Riy (reaction boundary condition for node i along y) in row 2i column e+ 2i.

(b) Store P ix (Load at node i along x axis) in vector P row 2i− 1, similarly, store P iy in row 2i.

5. Invert the matrix B, multiply it with the load vector P and solve for the unknown member forcesand reactions.

%% Initialize the statics matrix%b(1:2*npoin,1:2*npoin)=0.;%


Draft5–10 TRUSSES

% Determine direction cosines and insert them in b%for ielem=1:nelemnod1=lnods(ielem,1);nod2=lnods(ielem,2);lx=coord(nod2,1)-coord(nod1,1);ly=coord(nod2,2)-coord(nod1,2);l_elem(ielem)=alpha_i=...beta_i=...b(...,...)=...;b(...,...)=...;alpha_j=...beta_j=...b(...,...)=...b(...,...)=...

end%% Boundary conditions%nbc=0;for inode=1:npoinfor ibc=1:2if id(inode,ibc)==1

nbc=nbc+1;b(...,...)=1.;

endend

end

Example 5-2: Truss I, Matrix Method

Determine all member forces for the following truss

Solution:

1. We first determine the directions cosines



Member Nodes α β α β1 1-2 Node 1 Node 2

α11 = 1 β11 = 0 α12 = −1 β12 = 02 2-3 Node 2 Node 3

α22 = −.7071 β22 = .707 α23 = .707 β23 = −.7073 3-1 Node 3 Node 1

α33 = −.707 β33 = −.707 α31 = .707 β31 = .707

2. Next we write the equations of equilibrium

Node 1 ΣFx = 0ΣFy = 0



F1 F2 F3 R1x R1y R2y

α11 0 α31 1 0 0β11 0 β31 0 1 0α12 α22 0 0 0 0β12 β22 0 0 0 10 α23 α33 0 0 00 β23 β33 0 0 0

F1F2F3R1x

R1y

R2y

︸︷︷︸

{F}

+

00000−10

︸︷︷︸

{P}

= 0

Or




1 0 .707 1 0 00 0 .707 0 1 0−1 −.707 0 0 0 00 .707 0 0 0 10 .707 −.707 0 0 00 −.707 −.707 0 0 0

︸︷︷︸[B]

F1F2F3R1x

R1y

R2y

︸︷︷︸

{F}

+

00000−10

︸︷︷︸

{P}

= 0

Inverting [B] we obtain from F = [B]−1P

F1F2F3R1x

R1y

R2y

=

5−7.07−7.07

055

Example 5-3: Truss II, Matrix Method

Set up the statics matrix for the truss of Example 5-1 using the matrix methodSolution:First we number the joints and the elements as shown below.


Draft5–12 TRUSSES

Then we determine the direction cosines24√

242+322= 3

5 = 0.632√

242+322= 4

5 = 0.810√

242+102= 10

26 = 0.3824√

242+102= 24

26 = 0.92

# 1 ΣFxΣFy

# 2 ΣFxΣFy

# 3 ΣFxΣFy

# 4 ΣFxΣFy

# 5 ΣFxΣFy

# 6 ΣFxΣFy

# 7 ΣFxΣFy

# 8 ΣFxΣFy

1 .6 1.8 1

−1 11

−1 1 −.6 .6.8 1 .8

−1 11

−1 −.6.8 1

−.6 .6 −.92−.8 −1 −.8 .38

−.92 .92−1 −.38 −.38

−.6 .6 .92−.8 −1 −.8 .38

︸︷︷︸[B]

F1F2F3F4F5F6F7F8F9F10F11F12F13R1xR1yR5y

︸︷︷︸

?

+

000

−200

−400

−4000000

−80

−12

︸︷︷︸√

= 0

Using Mathematica we would have

b={

{ 1, 0., 0., 0.,0.6, 0., 0., 0., 0., 0., 0., 0., 0., 1, 0., 0.},

{0., 0., 0., 0.,0.8, 0., 0., 0., 0., 0., 0., 0., 0., 0., 1, 0.},

{-1, 1, 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.},

{0.,0., 0., 0., 0., 1, 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.},

{0.,-1, 1, 0., 0., 0.,-0.6, 0., 0.6, 0., 0., 0., 0., 0., 0., 0.},

{0., 0., 0., 0., 0., 0., 0.8, 1., 0.8, 0., 0., 0., 0., 0., 0., 0.},

{0., 0., -1, 1, 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.},

{0., 0., 0., 0., 0., 0., 0., 0., 0., 1., 0., 0., 0., 0., 0., 0.},

{0., 0., 0., -1, 0., 0., 0., 0., 0., 0.,-0.6, 0., 0., 0., 0., 0.},

{0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.8, 0., 0., 0., 0., 1.},

{0., 0., 0., 0., 0., 0., 0., 0., -0.6, 0., 0.6, 0.,-0.92, 0., 0., 0.},

{0., 0., 0., 0., 0., 0., 0., 0., -0.8,-1.,-0.8, 0., 0.38, 0., 0., 0.},

{0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,-0.92, 0.92, 0., 0., 0.},

{0., 0., 0., 0., 0., 0., 0., -1, 0., 0., 0.,-0.38,-0.38, 0., 0., 0.},

{0., 0., 0., 0., -0.6, 0., 0.6, 0., 0., 0., 0., 0.92, 0., 0., 0., 0.},

{0., 0., 0., 0.,-0.8, -1,-0.8, 0., 0., 0., 0., 0.38, 0., 0., 0., 0.}

}

p={0, 0, 0, -20, 0, -40 , 0, -40, 0, 0, 0, 0, 0, -8, 0, -12}

m=Inverse[b].p

And the result will be



Out[3]= {-43.5, -43.5, -46.5, -46.5, 72.5, -20., -7.66598, -31.7344,

-14

> -2.66598, -40., 77.5, 52.2822, 52.2822, -1.22125 10 , -58., -62.}

which correspond to the unknown element internal forces and external reactions.

5.2.3 Method of Sections

33 When only forces in selected members (away from loaded joints) is to be determined, this methodshould be used.

34 This method can be summarized as follows

1. “Cut” the truss into two substructures by an imaginary line (not necessarily straight) such that itwill at least intersect the member for which force is to be determined.

2. Consider either one of the two substructures as the free body

3. Each substructure must remain in equilibrium. Apply the equations of equilibrium

(a) Summation of moments about a particular point (usually the intersection of 2 cut members)would permit the determination of other member forces

(b) Summation of forces is usually used to determine forces in inclined members

Example 5-4: Truss, Method of Sections

Determine FBC and FHG in the previous example.Solution:Cutting through members HG, HC and BC, we first take the summation of forces with respect to H:


Draft5–14 TRUSSES

(+ �✁✛)ΣMH = 0 ⇒ RAy(24)− FBC(32) = 0

FBC = 2432 (58) = 43.5 k Tension

(+ �✁✛)ΣMC = 0; ⇒ (58)(24)(2)− (20 + 12)(24)− FHGx(32)− FHGy (24) = 0

2770− 768− (32)(FHG) 24√242+102

− (24)(FHG) 10√242+102

= 02, 000− (29.5)FHG − (9.2)FHG = 0

⇒ FHG = 52 k Compression

5.3 Case Study: Stadium

AB

D

F

C

E

G

7.5 m3 m 0.5 m

3 m

1 m

2 m

Figure 5.9: Florence Stadium, Pier Luigi Nervi (?)


Draft5.3 Case Study: Stadium 5–15

1

��

��

��

��

1 m

2 m

3 m

7.5 m

0.5 m

3 m

12

3

4

5 6

2 3

4

5

6

7

8

9

j=6

m=9

r=3

800 kN

Figure 5.10: Florence Stadioum, Pier Luigi Nervi (?)


Draft5–16 TRUSSES


Draft

Chapter 6

CABLES

6.1 Funicular Polygons

1 A cable is a slender flexible member with zero or negligible flexural stiffness, thus it can only transmittensile forces1.

2 The tensile force at any point acts in the direction of the tangent to the cable (as any other componentwill cause bending).

3 Its strength stems from its ability to undergo extensive changes in slope at the point of load application.

4 Cables resist vertical forces by undergoing sag (h) and thus developing tensile forces. The horizontalcomponent of this force (H) is called thrust.

5 The distance between the cable supports is called the chord.

6 The sag to span ratio is denoted by

r =h

l(6.1)

7 When a set of concentrated loads is applied to a cable of negligible weight, then the cable deflects intoa series of linear segments and the resulting shape is called the funicular polygon.

8 If a cable supports vertical forces only, then the horizontal componentH of the cable tension T remainsconstant.

Example 6-1: Funicular Cable Structure

Determine the reactions and the tensions for the cable structure shown below.

1Due to the zero flexural rigidity it will buckle under axial compressive forces.

Draft6–2 CABLES

Solution:We have 4 external reactions, however the horizontal ones are equal and we can use any one of a numberof equations of conditions in addition to the three equations of equilibrium. First, we solve for Ay,Dy and H = Ax = Dx. For this problem we could use the following 3 equations of static equilibriumΣFx = ΣFy = ΣM = 0, however since we do not have any force in the x direction, the second equationis of no avail. Instead we will consider the following set ΣFy = ΣMA = ΣMD = 0

1. First we solve for Dy

(+ �✁✛)ΣMA = 0;⇒ 12(30) + 6(70)−Dy(100) = 0⇒ Dy = 7.8 k (6.2)

2. Then we solve for Ay

(+ ✻)ΣFy = 0;⇒ Ay − 12− 6 + 7.8 = 0⇒ Ay = 10.2 k (6.3)

3. Solve for the horizontal force

(+ �✁✛)ΣMB = 0;⇒ Ay(30)−H(6) = 0⇒ H = 51 k (6.4)

4. Now we can solve for the sag at point C

(+ �✁✛)ΣMC = 0⇒ −DY (30) +H(hc) = 0⇒ hc =

30Dy

H=

30(7.8)51

= 4.6 ft (6.5)

5. We now solve for the cable internal forces or tractions in this case

TAB; tan θA =630

= 0.200⇒ θA = 11.31 deg (6.6-a)

=H

cos θA=

510.981

= 51.98 k (6.6-b)

TBC ; tan θB =6− 4.6

40= 0.035⇒ θB = 2deg (6.6-c)

=H

cos θB=

510.999

= 51.03 k (6.6-d)

TCD; tan θC =4.630

= 0.153⇒ θC = 8.7 deg (6.6-e)

=H

cos θC=

510.988

= 51.62 k (6.6-f)


Draft6.2 Uniform Load 6–3

6.2 Uniform Load

6.2.1 qdx; Parabola

9 Whereas the forces in a cable can be determined from statics alone, its configuration must be derivedfrom its deformation. Let us consider a cable with distributed load p(x) per unit horizontal projec-tion of the cable length2. An infinitesimal portion of that cable can be assumed to be a straight line,Fig. 6.1 and in the absence of any horizontal load we have H =constant. Summation of the vertical

H

T+dT

T

H

V

V+dV

H

θ

θ

q(x)

dx

dyds

ds

L

x

V

q(x)

y

y(x)

dx

h

y’

x’

x

y

L/2

Figure 6.1: Cable Structure Subjected to q(x)

forces yields

(+ ❄) ΣFy = 0⇒ −V + qdx+ (V + dV ) = 0 (6.7-a)dV + qdx = 0 (6.7-b)

where V is the vertical component of the cable tension at x3. Because the cable must be tangent to T ,we have

tan θ =V

H(6.8)

10 Substituting into Eq. 6.7-b yieldsd(H tan θ) + qdx = 0 (6.9)

or− ddx

(H tan θ) = q (6.10)

Since H is constant (no horizontal load is applied), this last equation can be rewritten as

−H d

dx(tan θ) = q (6.11)

11 Written in terms of the vertical displacement y, tan θ = dydx which when substituted in Eq. 6.11 yields

the governing equation for cables

d2y

dx2= − q

H(6.12)

2Thus neglecting the weight of the cable3Note that if the cable was subjected to its own weight then we would have qds instead of pdx.


Draft6–4 CABLES

12 For a cable subjected to a uniform load p. we can determine its shape by double integration of Eq.6.12

−Hy′ = qx+ C1 (6.13-a)

−Hy =qx2

2+ C1x+ C2 (6.13-b)

To solve for C1 and C2 this last equation must satisfy the boundary conditions: y = 0 at x = 0 and atx = L ⇒ C2 = 0 and C1 = − qL

2 . Thus

Hy =q

2x(L − x) (6.14)

13 This equation gives the shape y(x) in terms of the horizontal force H , it can be rewritten in terms ofthe maximum sag h which occurs at midspan, hence at x = L

2 we would have4

Hh =qL2

8(6.15)

14 This relation clearly shows that the horizontal force is inversely proportional to the sag h, as h ↘H ↗. Furthermore, this equation can be rewritten as

qL

H=

8hL

(6.16)

and combining this equation with Eq. 6.1 we obtain

qL

H= 8r (6.17)

15 Combining Eq. 6.14 and 6.15 we obtain

y =4hxL2

(L− x) (6.18)

16 If we shift the origin to midspan, and reverse y, then

y =4hL2x2 (6.19)

Thus the cable assumes a parabolic shape (as the moment diagram of the applied load).

17 The maximum tension occurs at the support where the vertical component is equal to V = qL2 and

the horizontal one to H , thus

Tmax =√V 2 +H2 =

√(qL

2

)2+H2 = H

√1 +

(qL/2H

)2(6.20)

Combining this with Eq. 6.17 we obtain5.

Tmax = H√

1 + 16r2 ≈ H(1 + 8r2) (6.21)

4Note the analogy between this equation and the maximum moment in a simply supported uniformly loaded beam

M = qL2

8.

5Recalling that (a + b)n = an + nan−1b +n(n−1)

2!an−2b2 + · or (1 + b)n = 1 + nb +

n(n−1)b2

2!+

n(n−1)(n−2)b3

3!+ · · ·;

Thus for b2 << 1,√1 + b = (1 + b)

12 ≈ 1 + b

2



6.2.2 † qds; Catenary

18 Let us consider now the case where the cable is subjected to its own weight (plus ice and wind if any).We would have to replace qdx by qds in Eq. 6.7-b

dV + qds = 0 (6.22)

The differential equation for this new case will be derived exactly as before, but we substitute qdx byqds, thus Eq. 6.12 becomes

d2y

dx2= − q

H

ds

dx(6.23)

19 But ds2 = dx2 + dy2, hence:

d2y

dx2= − q

H

√1 +

(dy

dx

)2(6.24)

solution of this differential equation is considerably more complicated than for a parabola.

20 We let dy/dx = p, thendp

dx= − q

H

√1 + p2 (6.25)

rearranging ∫dp√1 + p2

= −∫q

Hdx (6.26)

From Mathematica (or handbooks), the left hand side is equal to∫dp√1 + p2

= loge(p+√

1 + p2) (6.27)

Substituting, we obtain

loge(p+√

1 + p2) = −qxH

+ C1︸︷︷︸A

(6.28-a)

p+√

1 + p2 = eA (6.28-b)√1 + p2 = −p+ eA (6.28-c)1 + p2 = p2 − 2peA + e2A (6.28-d)

p =e2A − 1

2eA=eA − e−A

2= sinhA (6.28-e)

=dy

dx= sinh

(−qxH

+ C1)

(6.28-f)

y =∫

sinh(−qxH

+ C1)dx = −H

qcosh

(−qxH

+ C1)

+ C2 (6.28-g)

21 To determine the two constants, we set

dy

dx= 0 at x =

L

2(6.29-a)

dy

dx= − q

H

H

qsinh

(−qxH

+ C1)

(6.29-b)

⇒ 0 = sinh(− qH

L

2+ C1

)⇒ C1 =

q

H

L

2(6.29-c)

⇒ y = −Hq

cosh[q

H

(L

2− x)]

+ C2 (6.29-d)


Draft6–6 CABLES

At midspan, the sag is equal to h, thus

h = −Hq

cosh[q

H

(L

2− L

2

)]+ C2 (6.30-a)

C2 = h+H

q(6.30-b)

22 If we move the origin at the lowest point along the cable at x′ = x− L/2 and y′ = h− y, we obtain

q

Hy = cosh

( qHx)− 1 (6.31)

23 This equation is to be contrasted with 6.19, we can rewrite those two equations as:

q

Hy =

12

( qHx)2

Parabola (6.32-a)

q

Hy = cosh

( qHx)− 1 Catenary (6.32-b)

The hyperbolic cosine of the catenary can be expanded into a Taylor power series as

qy

H=

12

(qxH

)2+

124

(qxH

)4+

1720

(qxH

)6+ ... (6.33)

The first term of this development is identical as the formula for the parabola, and the other termsconstitute the difference between the two. The difference becomes significant only for large qx/H , thatis for large sags in comparison with the span, Fig. 6.3.

−2 −1.5 −1 −0.5 0 0.5 1 1.5 20

0.5

1

1.5

2

2.5

3

qx/H

ParabolaCatenary

Figure 6.2: Catenary versus Parabola Cable Structures

6.2.2.1 Historical Note

24 It should be mentioned that solution of this problem constitued one of the major mathematical/Mechanicschallenges of the early 18th century. Around 1684, differential and integral calculus took their first effec-tive forms, and those powerful new techniques allowed scientists to tackle complex problems for the firsttime, (Penvenuto 1991). One of these problems was the solution to the catenary problem as presented byJakob Bernouilli. Immediately thereafter, Leibniz presented a solution based on infinitesimal calculus,another one was presented by Huygens. Finally, the brother of the challenger, Johann Bernoulli did alsopresent a solution.



25 Huygens solution was complex and relied on geometrical arguments. The one of Leibniz was ellegantand correct (y/a = (bx/a+ b−x/a)/2 (we recognize Eq. 6.31 albeit written in slightly different form, Fig.6.3. Finally, Bernoulli presented two correct solution, and in his solution he did for the first time express

Figure 6.3: Leipniz’s Figure of a catenary, 1690

equations of equilibrium in differential form.

Example 6-2: Design of Suspension Bridge

Design the following 4 lanes suspension bridge by selecting the cable diameters assuming an allowablecable strength σall of 190 ksi. The bases of the tower are hinged in order to avoid large bending moments.

The total dead load is estimated at 200 psf. Assume a sag to span ratio of 15

Solution:

1. The dead load carried by each cable will be one half the total dead load or p1 = 12 (200) psf(50) ft 1

1,000 =5.0 k/ft


Draft6–8 CABLES

2. Using the HS 20 truck, the uniform additional load per cable isp2 = (2)lanes/cable(.64)k/ft/lane = 1.28 k/ft/cable. Thus, the total design load is p1 + p2 =5 + 1.28 = 6.28 k/ft

3. The thrust H is determined from Eq. 6.15

H =pl2

8h(6.34-a)

=(6.28) k/ft(300)2 ft2

(8)(3005 ft

) (6.34-b)

= 1, 177 k (6.34-c)

4. From Eq. 6.21 the maximum tension is

Tmax = H√

1 + 16r2 (6.35-a)

= (1, 177) k

√1 + (16)

(15

)2(6.35-b)

= 1, 507 k (6.35-c)

5. Note that if we used the approximate formula in Eq. 6.21 we would have obtained

Tmax = H(1 + 8r2) (6.36-a)

= 1, 177

(1 + 8

(15

)2)(6.36-b)

= 1, 554 k (6.36-c)

or 3% difference!

6. The required cross sectional area of the cable along the main span should be equal to

A =Tmaxσall

=1, 507 k

190 ksi= 7.93 in2

which corresponds to a diameter

d =

√4Aπ

=

√(4)(7.93)

π= 3.18 in

7. We seek next to determine the cable force in AB. Since the pylon can not take any horizontalforce, we should have the horizontal component of Tmax (H) equal and opposite to the horizontal

component of TAB or TABH =

√(100)2+(120)2

100 thus

TAB = H

√(100)2 + (120)2

100= (1, 177)(1.562) = 1, 838 k (6.37)

the cable area should beA =

1, 838 k

190 ksi= 9.68 in2

which corresponds to a diameter

d =

√(4)(9.68)

π= 3.51 in

8. To determine the vertical load acting on the pylon, we must add the vertical components of Tmaxand of TAB (VBC and VAB respectively). We can determine VBC from H and Tmax, thus

P =120100

(1, 177) +√

(1, 507)2 − (1, 177)2 = 1, 412 + 941 = 2, 353 k (6.38)

Using A36 steel with an allowable stress of 21 ksi, the cross sectional area of the tower should beA = 2,353

21 = 112 in2. Note that buckling of such a high tower might govern the final dimensions.


Draft6.3 Case Study: George Washington Bridge 6–9

9. If the cables were to be anchored to a concrete block, the volume of the block should be at leastequal to

V =(1, 412) k(1, 000)

150 lbs/ft3= 9, 413 ft3

or a cube of approximately 21 ft

6.3 Case Study: George Washington Bridge

Adapted from (Billington and Mark 1983)

26 The George Washington bridge, is a suspension bridge spanning the Hudson river from New YorkCity to New Jersey. It was completed in 1931 with a central span of 3,500 ft (at the time the world’slongest span). The bridge was designed by O.H. Amman, who had emigrated from Switzerland. In 1962the deck was stiffened with the addition of a lower deck.

6.3.1 Geometry

27 A longitudinal and plan elevation of the bridge is shown in Fig. 6.4. For simplicity we will assume in

??

377 ft 327 ft

610 ft 650 ft

ELEVATION

N.J. N.Y.HUDSON RIVER

PLAN

3,500 ft

4,760 ft

Figure 6.4: Longitudinal and Plan Elevation of the George Washington Bridge

our analysis that the two approaching spans are equal to 650 ft.

28 There are two cables of three feet diameter on each side of the bridge. The centers of each pair are 9ft apart, and the pairs themselves are 106 ft apart. We will assume a span width of 100 ft.

29 The cables are idealized as supported by rollers at the top of the towers, hence the horizontal com-ponents of the forces in each side of the cable must be equal (their vertical components will add up).

30 The cables support the road deck which is hung by suspenders attached at the cables. The cables aremade of 26,474 steel wires, each 0.196 inch in diameter. They are continuous over the tower supportsand are firmly anchored in both banks by huge blocks of concrete, the anchors.

31 Because the cables are much longer than they are thick (small IL), they can be idealized as perfectly

flexible members with no shear/bending resistance but with high axial strength.


Draft6–10 CABLES

32 The towers are 578 ft tall and rest on concrete caissons in the river. Because of our assumptionregarding the roller support for the cables, the towers will be subjected only to axial forces.

6.3.2 Loads

33 The dead load is composed of the weight of the deck and the cables and is estimated at 390 and 400psf respectively for the central and side spans respectively. Assuming an average width of 100 ft, thiswould be equivalent to

DL = (390) psf(100) ftk

(1, 000) lbs= 39 k/ft (6.39)

for the main span and 40 k/ft for the side ones.

34 For highway bridges, design loads are given by the AASHTO (Association of American State HighwayTransportation Officials). The HS-20 truck is often used for the design of bridges on main highways, Fig.6.5. Either the design truck with specified axle loads and spacing must be used or the equivalent uniformload and concentrated load. This loading must be placed such that maximum stresses are produced.

Figure 6.5: Truck Load

35 With two decks, we estimate that there is a total of 12 lanes or

LL = (12)Lanes(.64) k/ ft/Lane = 7.68 k/ft ≈ 8 k/ft (6.40)

We do not consider earthquake, or wind loads in this analysis.

36 Final DL and LL are, Fig. 6.6: TL = 39 + 8 = 47 k/ft

6.3.3 Cable Forces

37 The thrust H (which is the horizontal component of the cable force) is determined from Eq. 6.15

H =wL2cs8h

(6.41-a)

=(47) k/ft(3, 500)2 ft2

(8)(327) ft(6.41-b)

= 220, 000 k (6.41-c)



DEAD LOADS

Lw = 8 k/ft

D D,Sw = 39 k/ft w = 40 k/ftD,Sw = 40 k/ft

Figure 6.6: Dead and Live Loads

From Eq. 6.21 the maximum tension is

r =h

Lcs=

3273, 500

= 0.0934 (6.42-a)

Tmax = H√

1 + 16r2 (6.42-b)

= (220, 000) k√

1 + (16)(0.0934)2 (6.42-c)

= (220, 000) k(1.0675) = 235,000 k (6.42-d)

6.3.4 Reactions

38 Cable reactions are shown in Fig. 6.7.

POINTS WITH REACTIONS TOCABLES

Figure 6.7: Location of Cable Reactions

39 The vertical force in the columns due to the central span (cs) is simply the support reaction, 6.8

Vcs =12wLcs =

12(47) k/ft(3, 500) ft = 82, 250 k (6.43)


Draft6–12 CABLES

A

POINT OF NOMOMENT

B

REACTIONS ATTOP OF TOWER

TOT

L = 3,500 FT

w = 39 + 8 = 47 k/ft

Figure 6.8: Vertical Reactions in Columns Due to Central Span Load

Note that we can check this by determining the vector sum of H and V which should be equal to Tmax:√V 2cs +H

2 =√

(82, 250)2 + (220, 000)2 = 235, 000 k√

(6.44)

40 Along the side spans (ss), the total load is TL = 40+8 = 48 k/ft. We determine the vertical reactionby taking the summation of moments with respect to the anchor:

ΣMD = 0; �✁✛+;hssH + (wssLss)

Lss2− VssLss = 0 (6.45-a)

= (377) k(220, 000) k + (48) k/ft(650) ft(650) ft

2− 650Vss = 0 (6.45-b)

Vss = 143, 200 k (6.45-c)

41 Note: that we have used equilibrium to determine the vertical component of the cable force. It wouldhave been wrong to determine Vss from Vss = 220, 000 377650 as we did in the previous example, becausethe cable is now loaded. We would have to determine the shape of the cable and the tangent at thesupport. Beginiing with Eq. 6.13-b:

−Hy =12wx2 + C1x+ C2 (6.46-a)

−y =w

H

x2

2+C1Hx+

C2H

(6.46-b)

(6.46-c)

At x = 0, y = 0, thus C2 = 0; and at x = 650, y = −377; with H = 220, 000 k and w = 48

y = − (48) k

ft(220, 000) kx2 − C1

(220, 000) kx (6.47-a)

= −1.091× 10−4x2 − 4.545× 10−6C1x (6.47-b)y|x=650 = −377 = −1.091× 10−4(650)2 − 4.545× 10−6C1(650) (6.47-c)

C1 = 112, 000 (6.47-d)y = −1.091× 10−4x2 − 0.501x (6.47-e)dy

dx= −2.182× 10−4x− 0.501 (6.47-f)

dy

dx

∣∣∣∣x=650

= −0.1418− 0.501 = −0.6428 =V

H(6.47-g)

V = (220, 000)(0.6428) = 141, 423 k (6.47-h)

which is only 1% different.

42 Hence the total axial force applied on the column is

V = Vcs + Vss = (82, 250) k + (143, 200) k = 225,450 k (6.48)



43 The vertical reaction at the anchor is given by summation of the forces in the y direction, Fig. 6.9:

(+ ✻)ΣFy = 0; (wssLss) + Vss +Ranchor = 0 (6.49-a)−(48) k/ft(650) ft + (143, 200) k +Ranchor = 0 (6.49-b)

Ranchor = 112,000 k ❄ (6.49-c)

(6.49-d)

225,450 k

112,000 k

220,000 k

Figure 6.9: Cable Reactions in Side Span

44 The axial force in the side cable is determined the vector sum of the horizontal and vertical reactions.

T ssanchor =

√R2

anchor +H2 =√

(112, 000)2 + (220, 000)2 = 247, 000 k (6.50-a)

T sstower =

√V 2ss +H2 =

√(143, 200)2 + (220, 000)2 = 262,500 k (6.50-b)

45 The cable stresses are determined last, Fig. 6.10:

Awire =πD2

4=

(3.14)(0.196)2

4= 0.03017 in2 (6.51-a)

Atotal = (4)cables(26, 474)wires/cable(0.03017) in2/wire = 3, 200 in2(6.51-b)

Central Span σ =H

A=

(220, 000) k

(3, 200) in2= 68.75 ksi (6.51-c)

Side Span Tower σsstower =T sstowerA

=(262, 500) in2

(3, 200) in2= 82 ksi (6.51-d)

Side Span Anchor σsstower =T ssanchorA

=(247, 000) in2

(3, 200) in2= 77.2 ksi (6.51-e)

77.2 ksi

81.9 ksi73.4 ksi

68.75 ksi

Figure 6.10: Cable Stresses


Draft6–14 CABLES

46 If the cables were to be anchored to a concrete block, the volume of the block should be at least equalto V = (112,000) k(1,000) lbs/ k

150 lbs/ft3= 747, 000 ft3 or a cube of approximately 91 ft

47 The deck, for all practical purposes can be treated as a continuous beam supported by elastic springswith stiffness K = AL/E (where L is the length of the supporting cable). This is often idealized asa beam on elastic foundations, and the resulting shear and moment diagrams for this idealization areshown in Fig. 6.11.

K=AL/E

Shear

Moment

Figure 6.11: Deck Idealization, Shear and Moment Diagrams


Draft

Chapter 7

DESIGN PHILOSOPHIES of ACIand AISC CODES

7.1 Safety Provisions

1 Structures and structural members must always be designed to carry some reserve load above what isexpected under normal use. This is to account for

Variability in Resistance: The actual strengths (resistance) of structural elements will differ fromthose assumed by the designer due to:

1. Variability in the strength of the material (greater variability in concrete strength than insteel strength).

2. Differences between the actual dimensions and those specified (mostly in placement of steelrebars in R/C).

3. Effect of simplifying assumptions made in the derivation of certain formulas.

Variability in Loadings: All loadings are variable. There is a greater variation in the live loads thanin the dead loads. Some types of loadings are very difficult to quantify (wind, earthquakes).

Consequences of Failure: The consequence of a structural component failure must be carefully as-sessed. The collapse of a beam is likely to cause a localized failure. Alternatively the failure of acolumn is likely to trigger the failure of the whole structure. Alternatively, the failure of certaincomponents can be preceded by warnings (such as excessive deformation), whereas other are sud-den and catastrophic. Finally, if no redistribution of load is possible (as would be the case in astatically determinate structure), a higher safety factor must be adopted.

2 The purpose of safety provisions is to limit the probability of failure and yet permit economicalstructures.

3 The following items must be considered in determining safety provisions:

1. Seriousness of a failure, either to humans or goods.

2. Reliability of workmanship and inspection.

3. Expectation of overload and to what magnitude.

4. Importance of the member in the structure.

5. Chance of warning prior to failure.

4 Two major design philosophies have emerged

1. Working Stress Method

2. Ultimate Strength Method

Draft7–2 DESIGN PHILOSOPHIES of ACI and AISC CODES

7.2 Working Stress Method

5 This is the simplest of the two methods, and the one which has been historically used by structuralengineers.

6 Structural elements are designed for their service loads, and are dimensioned such that the stressesdo not exceed some predesignated allowable strength, Fig. 7.1.

Figure 7.1: Load Life of a Structure

7 In R/C this method was the one adopted by the ACI (American Concrete institute) code up to 1971,Working Stress Design Method (WSD).

8 The AISC (American Institute of Steel Construction) code refers to it as the Allowable StressDesign (ASD) and was used until 1986.

9 In this method:

1. All loads are assumed to have the same average variability.

2. The entire variation of the loads and the strengths is placed on the strength side of the equation.

σ < σall =σyldF.S.

(7.1)

where F.S. is the factor of safety.

10 Major limitations of this method

1. An elastic analysis can not easily account for creep and shrinkage of concrete.

2. For concrete structures, stresses are not linearly proportional to strain beyond 0.45f ′c.

3. Safety factors are not rigorously determined from a probabilistic approach, but are the result ofexperience and judgment.

11 Allowable strengths are given in Table 7.1.

7.3 Ultimate Strength Method

7.3.1 The Normal Distribution

12 The normal distribution has been found to be an excellent approximation to a large class of distribu-tions, and has some very desirable mathematical properties:


Draft7.3 Ultimate Strength Method 7–3

Steel, AISC/ASDTension, Gross Area Ft = 0.6FyTension, Effective Net Area∗ Ft = 0.5FuBending Fb = 0.66FyShear Fv = 0.40Fy

Concrete, ACI/WSDTension 0Compression 0.45f ′c

∗ Effective net area will be defined in section 9.2.1.2.

Table 7.1: Allowable Stresses for Steel and Concrete

1. f(x) is symmetric with respect to the mean µ.

2. f(x) is a “bell curve” with inflection points at x = µ± σ.3. f(x) is a valid probability distribution function as:∫ ∞

−∞f(x) = 1 (7.2)

4. The probability that xmin < x < xmax is given by:

P (xmin < x < xmax) =∫ xmax

xmin

f(x)dx (7.3)

13 The general normal (or Gauss) distribution is given by, Fig. 30.1:

−3.0 −2.5 −2.0 −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0x

0.00.10.20.30.40.50.60.70.80.91.0

CD

F

−3.0

−3.0

−2.5

−2.5

−2.0

−2.0

−1.5

−1.5

−1.0

−1.0

−0.5

−0.5

0.0

0.0

0.5

0.5

1.0

1.0

1.5

1.5

2.0

2.0

2.5

2.5

3.0

3.0

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

PD

F

2σ

µ

Figure 7.2: Normalized Gauss Distribution, and Cumulative Distribution Function

φ(x) =1√2πσ

e−12 [ x−µσ ]2 (7.4)



14 A normal distribution N(µ, σ2) can be normalized by defining

y =x− µσ

(7.5)

and y would have a distribution N(0, 1):

φ(y) =1√2πe−

y2

2 (7.6)

7.3.2 Reliability Index

15 In this approach, it is assumed that the load Q and the resistance R are random variables.

16 Typical frequency distributions of such random variables are shown in Fig. 7.3.

Figure 7.3: Frequency Distributions of Load Q and Resistance R

17 The safety margin is defined as Y = R−Q. Failure would occur if Y < 0

18 Q and R can be combined and the result expressed logarithmically, Fig. 7.4.

X = lnR

Q(7.7)

Failure would occur for negative values of X

19 The probability of failure Pf is equal to the ratio of the shaded area to the total area under thecurve in Fig. 7.4.

20 If X is assumed to follow a Normal Distribution than it has a mean value X =(ln R

Q

)m

and astandard deviation σ.

21 We define the safety index (or reliability index) as β = Xσ

22 For standard distributions and for β = 3.5, it can be shown that the probability of failure is Pf = 19,091

or 1.1 × 10−4. That is 1 in every 10,000 structural members designed with β = 3.5 will fail because ofeither excessive load or understrength sometime in its lifetime.

23 Reliability indices are a relative measure of the current condition and provide a qualitative estimateof the structural performance.

24 Structures with relatively high reliable indices will be expected to perform well. If the value is toolow, then the structure may be classified as a hazard.


Draft7.3 Ultimate Strength Method 7–5

Figure 7.4: Definition of Reliability Index

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5β

100

101

102

103

104

105

106

107

1/(P

roba

bilit

y of

Fai

lure

)

Probability of Failure in terms of β

Go

od

Haz

ard

ou

s

Un

sati

sfac

tory

Po

or

Bel

ow

Ave

rag

e

Ab

ove

Ave

rag

e

Figure 7.5: Probability of Failure in terms of β

25 Target values for β are shown in Table 30.2, and in Fig. 30.3

26 Because the strengths and the loads vary independently, it is desirable to have one factor to accountfor variability in resistance, and another one for the variability in loads.

27 These factors are referred to as resistance factor Φ and Load Factor α respectively. The resistancefactor is defined as

Φ =RmRn

exp(−0.55βVR) (7.8)

where RM RN and VR are the mean resistance, the nominal resistance (to be defined later), and thecoefficient of variation of the resistance.

7.3.3 Discussion

28 ACI refers to this method as the Strength Design Method, (previously referred to as the UltimateStrength Method).



Type of Load/Member βAISC

DL + LL; Members 3.0DL + LL; Connections 4.5DL + LL + WL; Members 3.5DL + LL +EL; Members 1.75

ACIDuctile Failure 3-3.5Sudden Failures 3.5-4

Table 7.2: Selected β values for Steel and Concrete Structures

29 AISC refers to it as Load and Resistance Factor Design (LRFD).

30 Terms such as failure load should be avoided; it is preferable to refer to a structure’s Limit Stateload.

31 The general form is (LRFD-A4.1)1

ΦRn ≥ ΣαiQi (7.9)

where

Φ is a strength reduction factor, less than 1, and must account for the type of structural element,Table 7.3.

Type of Member ΦACI

Axial Tension 0.9Flexure 0.9Axial Compression, spiral reinforcement 0.75Axial Compression, other 0.70Shear and Torsion 0.85Bearing on concrete 0.70

AISCTension, yielding 0.9Tension, fracture 0.75Compression 0.85Beams 0.9Fasteners, Tension 0.75Fasteners, Shear 0.65

Table 7.3: Strength Reduction Factors, Φ

Rn is the nominal resistance (or strength).

ΦRn is the design strength.

αi is the load factor corresponding to Qi and is greater than 1.

ΣαiQi is the required strength based on the factored load:

i is the type of load

32 The various factored load combinations which must be considered are1Throughout the notes we will refer by this symbol the relevant design specification in the AISC code.


Draft7.4 Example 7–7

AISC

1. 1.4D

2. 1.2D+1.6L+0.5(Lr or S)

3. 1.2D+0.5L (or 0.8W)+1.6(Lr or S)

4. 1.2D+0.5L+0.5(Lr or S)+1.3W

5. 1.2D+0.5L(or 0.2 S)+1.5E

6. 0.9D+1.3W(or 1.5 E)

ACI

1. 1.4D+1.7L

2. 0.75(1.4D+1.7L+1.7W)

3. 0.9D+1.3W

4. 1.05D+1.275W

5. 0.9D+1.7H

6. 1.4D +1.7L+1.7H

7. 0.75(1.4D+1.4T+1.7L)

8. 1.4(D+T)

where D= dead; L= live; Lr= roof live; W= wind; E= earthquake; S= snow; T= temperature; H= soil.We must select the one with the largest limit state load.

33 Thus, in this method, we must perform numerous analysis, one for each load, of a given structure.For trusses, this is best achieved if we use the matrix method, invert the statics matrix [B], and multiply[B]−1 by each one of the load cases, (Refer to Section 5.2.2.1). For the WSD method, we need notperform more than one analysis in general.

34 Serviceability Limit States must be assessed under service loads (not factored). The most impor-tant ones being

1. Deflections

2. Crack width (for R/C)

3. Stability

35 Most code requirements on strength are written as:

ΦxRn > Qu (7.10)

where Rn is the nominal strength of the member, Qu is the ultimate (factored) load effect (force, moment,shear, torque). Design load is also at times defined as

Rd =QuΦx

(7.11)

7.4 Example

Example 7-1: LRFD vs ASD



To illustrate the differences between the two design approaches, let us consider the design of an axialmember, subjected to a dead load of 100 k and live load of 80 k. Use A36 steel.

ASD: We consider the total load P = 100 + 80 = 180 k. From Table 7.1, the allowable stress is0.6σyld = 0.6 ∗ 36 = 21.6 ksi. Thus the required cross sectional area is

A =18021.6

= 8.33 in2

USD we consider the largest of the two load combinations

ΣαiQi : 1.4D = 1.4(100) = 140 k

1.2D + 1.6L = 1.2(100) + 1.6(80) = 248 k ✛

From Table 7.3 Φ = 0.9, and ΦRn = (0.9)Aσyld. Hence, applying Eq. 7.9 the cross sectional areashould be

A =ΣαiQiΦσyld

=248

(0.9)(36)= 7.65 in2

Note that whereas in this particular case the USD design required a smaller area, this may not be thecase for different ratios of dead to live loads.


Draft

Chapter 8

DESIGN I

8.1 Case Study: Eiffel Tower


8.1.1 Materials, & Geometry

1 The tower was built out of wrought iron, less expensive than steel,and Eiffel had more experience withthis material, Fig. 8.1

Figure 8.1: Eiffel Tower (Billington and Mark 1983)

2 The structure is essentially a tower, subjected to gravity and wind load. It is a relatively “light”structure, so dead load is small compared to the wind load.

3 To avoid overturning MgravityMwind had to be much higher than 1. This can be achieved either by:

1. Increase self weight (as in Washington’s monument)

2. Increase the width of the base

Draft8–2 DESIGN I

3. Design support to resist tension.

4. Post-tension the support.

4 The tower is 984 feet high, and 328 feet wide at the base. The large base was essential to provideadequate stability in the presence of wind load.

5 We can assume that the shape of the tower is parabolic. If we take the x axis to be along the verticalaxis of symmetry and y the half width, then we know that at x = 984 the (half) width y = 0 and atx = 0 the half width is 328/2 = 164, thus the equation of the half width is given by

y = 164(

984− x984

)2︸︷︷︸

av(x)2

(8.1)

6 We recall from calculus that for y = v(x)

dy

dx=

dy

dv

dv

dx(8.2-a)

d

dxax2 = 2ax (8.2-b)

Thus for our problem

dy

dx= 2(164)

984− x984︸︷︷︸

dydv

(− 1

984

)︸︷︷︸

dvdx

(8.3-a)

=984− x2, 952

(8.3-b)

Alsody

dx= tanβ ⇒ β = tan−1 dy

dx(8.4)

where β is the angle measured from the x axis to the tangent to the curve.

7 We now can tabulate the width and slope at various elevations

WidthLocation Height Width/2 Estimated Actual dy

dx βSupport 0 164 328 .333 18.4o

First platform 186 108 216 240 .270 15.1o

second platform 380 62 123 110 .205 11.6o

Intermediate platform 644 20 40 .115 6.6o

Top platform 906 1 2 .0264 1.5o

Top 984 0 0 0.000 0o

8 The tower is supported by four inclined supports, each with a cross section of 800 in2. An idealizationof the tower is shown in Fig. 8.2.

8.1.2 Loads

9 The total weight of the tower is 18, 800 k.

10 The dead load is not uniformly distributed, and is approximated as follows, Fig. 8.3:

Location Height Dead WeightGround- second platform 380 ft 15, 500 k

Second platform-intermediate platform 264 ft 2, 200 k

intermediate platform - top 340 ft 1, 100 k

Total 984 ft 18, 800 kVictor Saouma Structural Engineering

Draft8.1 Case Study: Eiffel Tower 8–3

ACTUALCONTINUOUSCONNECTION

ACTUALPOINTS OFCONNECTION

CONTINUOUSCONNECTION

IDEALIZED

Figure 8.2: Eiffel Tower Idealization, (Billington and Mark 1983)

Figure 8.3: Eiffel Tower, Dead Load Idealization; (Billington and Mark 1983)


Draft8–4 DESIGN I

11 From the actual width of the lower two platforms we can estimate the live loads (the intermediateand top platforms would have negligible LL in comparison):

1st platform: (50) psf(240)2 ft2 kip

(1,000) lbs 2, 880 k

2nd platform: (50) psf(110)2 ft2 k(1,000) lbs 600 k

Total: 3, 480 k

Hence the total vertical load is Pvert = DL+ LL = 18, 800 + 3, 480 = 22, 280 k.

12 The wind pressure is known to also have a parabolic distribution (maximum at the top), the crosssectional area over which the wind is acting is also parabolic (maximum at the base). Hence we willsimplify our analysis by considering an equivalent wind force obtained from a constant wind pressure(force/length) and constant cross section Fig. 8.4: The pressure is assumed to be 2.6 k/ft, thus the

Figure 8.4: Eiffel Tower, Wind Load Idealization; (Billington and Mark 1983)

lateral wind force is, Fig. 8.5

P=2560k

Q=22,280kL/2

TOTALLOADS

��

��

��

��

H

V

M 0

Q

P

0

0

LOADS

REACTIONS

Figure 8.5: Eiffel Tower, Wind Loads, (Billington and Mark 1983)

Plat = (2.6) k/ft(984) ft = 2,560 k acting at9842

= 492 ft (8.5)



8.1.3 Reactions

13 Simplifying the three dimensional structure with 4 supports into a two dimensional one with twosupports, the reactions can be easily determined for this statically determinate structure, Fig. 8.6.

VERTICALFORCES FORCES

TOTAL

WINDWARDSIDE

LEEWARD

SIDE

+ =

WIND

Figure 8.6: Eiffel Tower, Reactions; (Billington and Mark 1983)

Gravity Load

Pvert = 22, 280 ❄ (8.6-a)

Rgravvert =

22, 2802

= 11,140 k ✻ (8.6-b)

Lateral Load

Lateral Moment (we essentially have a cantilevered beam subjected to a uniform load). Themoment at a distance x from the support along the cantilevered beam subjected to a uniformpressure p is given by

Mlat = p(L− x)︸︷︷︸Force

L− x2︸︷︷︸

Moment arm

= p(L− x)2

2(8.7)

Thus the lateral moment caused by the wind is parabolic. At the base (x = 0), the maximummoment is equal to

Mlat = p(L− x)2

2= (2.6) k/ft

(984− 0)2

2ft2 = 1,260,000 k.ft

�✁✛ (8.8)

We observe that the shape of the moment diagram is also parabolic, just like the tower itself.This is not accidental, as nearly optimum structures have a shape which closely approximatetheir moment diagram (such as the varying depth of continuous long span bridges).To determine the resulting internal forces caused by the lateral (wind) moment, and since wehave two supports (one under tension and the other under compression) we use

Rwindvert = ±M

d=

1, 260, 000 k.ft

328 ft= ± 3,850 k ✻❄ (8.9)

Lateral Forces to be resisted by each of the two pairs. By symmetry, the lateral force will beequally divided among the two pairs of supports and will be equal to

Rwindlat =

(2, 560) k

2= 1,280 k ✛ (8.10)


Draft8–6 DESIGN I

8.1.4 Internal Forces

14 First, a brief reminder

F

F F

x

y

θ

θ cos θ =FxF

(8.11-a)

sin θ =FyF

(8.11-b)

tan θ =FyFx

(8.11-c)

15 Internal forces are first determined at the base.

16 Gravity load are first considered, remember those are caused by the dead load and the live load,Fig. 8.7:

INCLINEDINTERNALFORCE: N

KNOWN VERTICALCOMPONENT: VCOMPONENT: H

HORIZONTALCONSEQUENT

β=18.40

V

H

N

β=18.4

FORCE POLYGON

Figure 8.7: Eiffel Tower, Internal Gravity Forces; (Billington and Mark 1983)

cosβ =V

N⇒ N =

V

cosβ(8.12-a)

N =11, 140 k

cos 18.4o= 11,730 kip (8.12-b)

tanβ =H

V⇒ H = V tanβ (8.12-c)

H = 11, 140 k(tan 18.4o) = 3,700 kip (8.12-d)

The horizontal forces which must be resisted by the foundations, Fig. 8.8.

H H

3700 k 3700 k

Figure 8.8: Eiffel Tower, Horizontal Reactions; (Billington and Mark 1983)

17 Because the vertical load decreases with height, the axial force will also decrease with height.

18 At the second platform, the total vertical load is Q = 1, 100+ 2, 200 = 3, 300 k and at that height the



angle is 11.6o thus the axial force (per pair of columns) will be

Nvert =3,300 k

2

cos 11.6o= 1, 685 k (8.13-a)

Hvert =3, 300 k

2(tan 11.6o) = 339 k (8.13-b)

Note that this is about seven times smaller than the axial force at the base, which for a given axialstrength, would lead the designer to reduce (or taper) the cross-section.

The horizontal force will be resisted by the axial forces in the second platform itself.

19 Wind Load: We now have determined at each pair of support the vertical and the horizontal forcescaused by the wind load, the next step is to determine their axial components along the inclined leg,Fig. 8.9:

1,280 k

1,280 sin 18.4

3,850 cos 18.43,850 k

18.4

18.4Figure 8.9: Eiffel Tower, Internal Wind Forces; (Billington and Mark 1983)

Nc = −Rwindvert cosβ −Rwind

lat sinβ (8.14-a)= −(3, 850) k(cos 18.4o)− (1, 280) k(sin 18.40) (8.14-b)

= -4,050 k Leeward (8.14-c)

Nt = −Rwindvert cosβ +Rwind

lat sinβ (8.14-d)= (3, 850) k(cos 18.4o) + (1, 280) k(sin 18.40) (8.14-e)

= 4,050 k Winward (8.14-f)

8.1.5 Internal Stresses

20 The total forces caused by both lateral and gravity forces can now be determined:

NTotalL = −(11, 730) k︸︷︷︸

gravity

−(4, 050) k︸︷︷︸lateral

= -15,780 k Leeward side (8.15-a)

NTotalW = −(11, 730) k︸︷︷︸

gravity

+(4, 050) k︸︷︷︸lateral

= -7,630 k Winward side (8.15-b)

We observe that even under wind load, the windward side is still under compression.

21 In the idealization of the tower’s geometry, the area of each pair of the simplified columns is 1, 600 in2.and thus the maximum stresses will be determined from

σcomp =NTL

A=−15, 780 k

1, 600 in2= -9.9 ksi (8.16)

22 The strength of wrought iron is 45 ksi, hence the safety factor is

Safety Factor =ultimate stressactual stress

=45 ksi

9.9 ksi= 4.5 (8.17)


Draft8–8 DESIGN I

8.2 Magazini Generali by Maillart


8.2.1 Geometry

23 This storage house, built by Maillart in Chiasso in 1924, provides a good example of the marriagebetween aesthetic and engineering.

24 The most striking feature of the Magazini Generali is not the structure itself, but rather the shape ofits internal supporting frames, Fig. 8.10.

HINGE IDEALIZATION

OF THIN SECTIONS

ACTUAL FRAME

AS A SIMPLE BEAMABSTRACTION OF MID SECTION

63.6 ft

9.2 ft

Figure 8.10: Magazzini Generali; Overall Dimensions, (Billington and Mark 1983)

25 The frame can be idealized as a simply supported beam hung from two cantilever column supports.Whereas the beam itself is a simple structural idealization, the overhang is designed in such a way as tominimize the net moment to be transmitted to the supports (foundations), Fig. 8.11.

8.2.2 Loads

26 The load applied on the frame is from the weights of the roof slab, and the frame itself. Given thespace between adjacent frames is 14.7 ft, and that the roof load is 98 psf, and that the total frameweight is 13.6 kips, the total uniform load becomes, Fig. 8.12:

qroof = (98) psf(14.7) ft = 1.4 k/ft (8.18-a)

qframe =(13.6) k

(63.6) ft= 0.2 k/ft (8.18-b)

qtotal = 1.4 + 0.2 = 1.6 k/ft (8.18-c)


Draft8.2 Magazini Generali by Maillart 8–9

d d

M =P*d

P

12

2P

M =B*dB 1

M =M -MR B P

B

B

Figure 8.11: Magazzini Generali; Support System, (Billington and Mark 1983)

ROOFq = 1.4 k/ft

FRAME + q = 0.2 k/ft

ROOFq = 1.4 k/ft FRAME

q = 1.6 k/ftTOTAL

+ q = 0.2 k/ft

Figure 8.12: Magazzini Generali; Loads (Billington and Mark 1983)

8.2.3 Reactions

27 Reactions for the beam are determined first taking advantage of symmetry, Fig. 8.13:

W = (1.6) k/ft(63.6) ft = 102 k (8.19-a)

R =W

2=

1022

= 51 k (8.19-b)

We note that these reactions are provided by the internal shear forces.

63.6 ft

51 k 51 k

TOTALq = 1.6 k/ft

Figure 8.13: Magazzini Generali; Beam Reactions, (Billington and Mark 1983)


Draft8–10 DESIGN I

8.2.4 Forces

28 The internal forces are primarily the shear and moments. Those can be easily determined for asimply supported uniformly loaded beam. The shear varies linearly from 51 kip to -51 kip with zero atthe center, and the moment diagram is parabolic with the maximum moment at the center, Fig. 8.14,equal to:

Mmax =qL2

8=

(1.6) k/ft(63.6) ft2

8= 808 k.ft (8.20)

Mmax

SHE

AR

FO

RC

E

51 K

25 K

25 K

51 K

0L/2

Lx

x

MO

ME

NT

L

0 L/4 L/2 3L/4

Figure 8.14: Magazzini Generali; Shear and Moment Diagrams (Billington and Mark 1983)

29 The externally induced moment at midspan must be resisted by an equal and opposite internalmoment. This can be achieved through a combination of compressive force on the upper fibers, andtensile ones on the lower. Thus the net axial force is zero, however there is a net internal couple, Fig.8.15.

A

TOTALq

VA

C

T

d M

Figure 8.15: Magazzini Generali; Internal Moment, (Billington and Mark 1983)

Mext = Cd⇒ C =Mext

d(8.21-a)

T = C =(808) k.ft

(9.2) ft= ± 88 k (8.21-b)


Draft8.2 Magazini Generali by Maillart 8–11

30 Because the frame shape (and thus d(x)) is approximately parabolic, and the moment is also parabolic,then the axial forces are constants along the entire frame, Fig. 8.16.

MOMENT DIAGRAM

M

CURVE OF DIAGRAM

FRAME :CABLE :

SHAPE OF DIAGRAM

FRAME

d

Figure 8.16: Magazzini Generali; Similarities Between The Frame Shape and its Moment Diagram,(Billington and Mark 1983)

31 The axial force at the end of the beam is not balanced, and the 88 kip compression must be transmittedto the lower chord, Fig. 8.17. Fig. 8.18 This is analogous to the forces transmitted to the support by a

Tied ArchCable Force

Vertical Reaction

Horizontal Component

Axial Force

Compression

88 k

88 kTension88 k

88 k

Figure 8.17: Magazzini Generali; Equilibrium of Forces at the Beam Support, (Billington and Mark1983)

tied arch.

32 It should be mentioned that when a rigorous computer analysis was performed, it was determinedthat the supports are contributing a compression force of about 8 kips which needs to be superimposedover the central values, Fig. 8.18.


33 The net compressive stress, for a top chord with a cross sectional area of 75 in2 is equal to

σ =P

A=

(88) k

(75) in2= 1.17 ksi (8.22)

this is much lower than the allowable compressive stress of concrete which is about 1,350 ksi. It shouldbe noted that if the frame was cast along with the roof (monolithic construction), than this stress wouldbe even lower.



FRAME ACTS AS A

UNIT, UNLIKE THE

ABSTRACTION

16 k

16 k

88 k - 8 k = 80 k

-88 k - 8 k = -96 k

Figure 8.18: Magazzini Generali; Effect of Lateral Supports, (Billington and Mark 1983)

34 Since concrete has practically no tensile strength, the tensile force in the lower chord must be resistedby steel. The lower chord has 4 bars with 0.69 in2 and 6 other bars with 0.58 in2, thus we have a totalof

As = 4(0.69) + 6(0.58) = 6.24 in2 (8.23)

Thus the steel stresses will be

σ =P

A=

(88) k

(6.24) in2= 14.1 ksi (8.24)

which is lower than the allowable steel stress.

8.3 Buildings

8.3.1 Buildings Structures

35 There are three primary types of building systems:

Wall Subsystem: in which very rigid walls made up of solid masonry, paneled or braced timber, orsteel trusses constitute a rigid subsystem. This is only adequate for small rise buildings.

Vertical Shafts: made up of four solid or trussed walls forming a tubular space structure. The tubularstructure may be interior (housing elevators, staircases) and/or exterior. Most efficient for veryhigh rise buildings.

Rigid Frame: which consists of linear vertical components (columns) rigidly connected to stiff hor-izontal ones (beams and girders). This is not a very efficient structural form to resist lateral(wind/earthquake) loads.

8.3.1.1 Wall Subsystems

36 Whereas exterior wall provide enclosure and interior ones separation, both of them can also have astructural role in transferring vertical and horizontal loads.


Draft8.3 Buildings 8–13

37 Walls are constructed out of masonry, timber concrete or steel.

38 If the wall is braced by floors, then it can provide an excellent resistance to horizontal load in theplane of the wall (but not orthogonal to it).

39 When shear-walls subsystem are used, it is best if the center of orthogonal shear resistance is closeto the centroid of lateral loads as applied. If this is not the case, then there will be torsional designproblems.

8.3.1.1.1 Example: Concrete Shear Wall From (Lin and Stotesbury 1981)

40 We consider a reinforced concrete wall 20 ft wide, 1 ft thick, and 120 ft high with a vertical load of400 k acting on it at the base. As a result of wind, we assume a uniform horizontal force of 0.8 kip/ftof vertical height acting on the wall. It is required to compute the flexural stresses and the shearingstresses in the wall to resist the wind load, Fig. 8.19.

��

��

��

+ F

H=96 k;M =5760 k’

120’

1’

- f

- F

+ F

HORIZONTAL

VERTICAL

2/(3d)

7.7’ IN TENSION

+140

+ 600+ 740 PSI

+ f

V

60’

DL

M M

20’

400 kW

w=0.8 k/ft

Figure 8.19: Design of a Shear Wall Subsystem, (Lin and Stotesbury 1981)

1. Maximum shear force and bending moment at the base

Vmax = wL = (0.8) k.ft(120) ft = 96 k (8.25-a)

Mmax =wL2

2=

(0.8) k.ft(120)2 ft2

2= 5, 760 k.ft (8.25-b)

2. The resulting eccentricity is

eActual =M

P=

(5, 760) k.ft

(400) k= 14.4 ft (8.26)

3. The critical eccentricity is

ecr =L

6=

(20) ft

6= 3.3 ft < eActualN.G. (8.27)

thus there will be tension at the base.



4. The moment of inertia of the wall is

I =bh3

12=

(1) ft(20) ft3

12= 667 ft4 (8.28)

5. The maximum flexural stresses will be

σmax = ±McI

=(5, 760) k.ft(10) ft

(667) ft4= ±(86.5) ksf = ±(600) psi (8.29)

6. The average shearing stress is

τ =V

A=

(96) k

(1)(20) ft2= 4.8 ksf = 33.3 psi (8.30)

A concrete with nominal shear reinforcement can carry at least 100 psi in shear, those computedshear stresses are permissible.

7. At the base of the wall, the axial stresses will be

σ =−(400) k

(1)(20) ft2= (20) ksf =≈ −140 psi (8.31-a)

8. The maximum stresses will thus be:

σ1 = −140 + 600 = 460 psi (Tension) (8.32-a)

σ2 = −140− 600 = −740 psi (Compression) (8.32-b)

9. The compressive stress of 740 psi can easily be sustained by concrete, as to the tensile stress of460 psi, it would have to be resisted by some steel reinforcement.

10. Given that those stresses are service stresses and not factored ones, we adopt the WSD approach,and use an allowable stress of 20 ksi, which in turn will be increased by 4/3 for seismic and windload,

σall =43(20) = 26.7 ksi (8.33)

11. The stress distribution is linear, compression at one end, and tension at the other. The length ofthe tension area is given by (similar triangles)

x

460=

20460 + 740

⇒ x =460

460 + 740(20) = 7.7 ft (8.34)

12. The total tensile force inside this triangular stress block is

T =12(460) ksi(7.7× 12) in (12) in︸︷︷︸

width

= 250 k (8.35)

13. The total amount of steel reinforcement needed is

As =(250) k

(26.7) ksi= 9.4 in2 (8.36)

This amount of reinforcement should be provided at both ends of the wall since the wind orearthquake can act in any direction. In addition, the foundations should be designed to resisttensile uplift forces (possibly using piles).


Draft8.3 Buildings 8–15

20’

120’

24’

1.2

1

~1.6

V

60’

W400 k

+F -FMM

H=96 k

Figure 8.20: Trussed Shear Wall

8.3.1.1.2 Example: Trussed Shear Wall From (Lin and Stotesbury 1981)

41 We consider the same problem previously analyzed, but use a trussed shear wall instead of a concreteone, Fig. 8.20.

1. Using the maximum moment of 5, 760 kip-ft (Eq. 8.25-b), we can compute the compression andtension in the columns for a lever arm of 20 ft.

F = ± (5, 760) k.ft

(20) ft= ±288 k (8.37)

2. If we now add the effect of the 400 kip vertical load, the forces would be

C = − (400) k

2− 288 = −488 k (8.38-a)

T = − (400) k

2+ 288 = 88 k (8.38-b)

3. The force in the diagonal which must resist a base shear of 96 kip is (similar triangles)

F

96=

√(20)2 + (24)2

20⇒ F =

√(20)2 + (24)2

20(96) = 154 k (8.39)

4. The design could be modified to have no tensile forces in the columns by increasing the width ofthe base (currently at 20 ft).

8.3.1.2 Shaft Systems

42 Vertical shear-resisting shafts in buildings act as a tubular section and generally have a rectangularcross section. If there is only one shaft, it is generally located in the center and houses the elevators. Ifthere are many shafts, then they should be symmetrically arranged.

43 If the shaft is relatively short and wide, with an aspect ratio under 1 or 2, then the dominant structuralaction is that of a stiff shear resisting tube. If the aspect ratio is between 3 and 5, then the shear forcesmay not be the controlling criterion, and flexure dominates.



8.3.1.2.1 Example: Tube Subsystem From (Lin and Stotesbury 1981)

44 With reference to Fig. 8.21, the reinforced concrete shaft is 20 ft square, 120 ft high, and with 1 ft

~ 20 ’~ 20 ’

N.A.

��

��

��

��

w = 0.8 k/ft

H = 96 k

60 ’

120 ’

20 ’ 20 ’

Figure 8.21: Design Example of a Tubular Structure, (Lin and Stotesbury 1981)

thick walls. It is subjected to a lateral force of 0.8 k/ft.

1. Comparing this structure with the one analyzed in Sect. 8.3.1.1.1 the total vertical load acting onthe base is now increased to

V = 4(400) = 1, 600 k (8.40)

2. As previously, the maximum moment and shear are 5, 760 k.ft and 96 k respectively.

3. The moment of inertia for a tubular section is

I = Σbd3

12=

(20)(20)3

12− (18)(18)3

12= 4, 600 ft4 (8.41)

4. The maximum flexural stresses:

σfl = ±MCI

= ± (5, 760) k.ft(20/2) ft

(4, 600) ft4= ±12.5 ksf = ±87 psi (8.42)

5. The average shear stress is

τ =V

A=

(96) k

2(20)(1) ft2= 2.4 ksf = 17 psi (8.43)

6. The vertical load of 1,600 k produces an axial stress of

σax =P

A=−(1, 600) k

(4(20)(1) ft2= −20 ksf = −140 psi (8.44)

7. The total stresses are thus

σ = σax + σfl (8.45-a)

σ1 = −140 + 87 = −53 psi (8.45-b)

σ2 = −140− 87 = −227 psi (8.45-c)

thus we do not have any tensile stresses, and those stresses are much better than those obtainedfrom a single shear wall.


Draft8.4 Design of a Three Hinged Arch 8–17

8.3.1.3 Rigid Frames

45 Rigid frames can carry both vertical and horizontal loads, however their analysis is more complexthan for tubes.

46 The rigorous and exact analysis of a rigid frame can only be accomplished through a computer analysis.However, for preliminary design it is often sufficient to perform approximate analysis.

47 There are two approximate methods for the analysis of rigid frames subjected to lateral loads: 1)Portal and 2) Cantilever method.

48 The portal frame method is based on the following major assumptions, Fig. 8.22:

L

P

h

h/2

L/2

PI

P P

h/2

V =P/(2L) V =P/(2L)

H =P/2H =P/2

1

1 2

2

Figure 8.22: A Basic Portal Frame, (Lin and Stotesbury 1981)

1. Each bay of a bent acts as a separate “portal” frame consisting of two adjacent columns and theconnecting girder.

2. The point of inflection (zero moment) for all columns is at midheight

3. The point of inflection for all girders is at midspan.

4. For a multibay frame, the shears on the interior columns are equal and the shear in each exteriorcolumn is half the shear of an interior column.

8.4 Design of a Three Hinged Arch

adapted from (Lin and Stotesbury 1981)A long arch 100 ft high and spanning 510 ft is to be designed for a garage and hotel building, using

air rights over roads and highways.

510’

100’Garage and hotel Building

It is necessary to determine preliminary dimensions for the size of the arch section. The arches arespaced 60 ft on centers and carry four-story loading totaling 27 k/ft along each arch.

1. To the initial DL and LL of 27 k/ft we add the arch own weight estimated to be ≈ 25% of theload, thus the total load is

w = (1 + .25)27 = 33.7 ≈ 33 k/ft (8.46)



2. We next determine the various forces:

H =wL2

8h=

(33)(510)2

8(100)= 10, 700 k (8.47-a)

V =wL

2=

(33)(510)2

= 8, 400 k (8.47-b)

R =√H2 + V 2 =

√(10, 700)2 + (8, 400)2 = 13, 600 k (8.47-c)

3. If we use concrete-filled steel pipe for arch section, and selecting a pipe diameter of 6 ft with athickness of 1/2 inch, then the steel cross sectional area is

As = 2πrt = πDt = (3.14)(6) ft(12) in/ ft(0.5) in = 113 in2 (8.48)

4. The concrete area is

Ac = πD2

4= (3.14)

(6)2

4ft2(144) in2/ftsq = 4, 070 in2 (8.49)

5. Assuming that the steel has an allowable stress of 20 ksi and the concrete 2.5 ksi (noting that thestrength of confined concrete can be as high as three times the one of f ′c), then the load carry-

ing capacity of each component isSteel Ac (113)(20)ksi = 2,260 k

Concrete As (4, 070)(2.5)ksi = 10,180 kTotal 12,440 kip

which is o.k. for the crown section (H=10,700 k) but not quite for the abutments at R=13,600 k.

6. This process of trial and error can be repeated until a satisfactory preliminary design is achieved.Furthermore, a new estimate for the arch self weight should be undertaken.

8.5 Salginatobel Bridge (Maillart)


8.5.1 Geometry

49 The Salginatobel bridge, perhaps the most famous and influential structure of Maillart is located inhigh up in the Swiss Alps close to Shuders.

50 It is a three hinged pedestrian bridge which crosses a deep valley with a most beautiful shape whichblends perfectly with its surrounding, Fig. 8.23

51 The load supporting structure is the arch itself, whereas the bridge deck and the piers are transferringthe vertical load into the arch.

52 The arch cross section is not constant, and can be idealized as in Fig. 8.24

53 The basic shape of the supporting structure is a three hinged arch as shown in Fig. 8.25

54 The arch is parabolic (which as we saw an the optimal shape which minimizes flexure), and the crosssection at the quarter point has an area of

At = 2[(0.62)(12.46) + (0.59)(12.17)] = 29.8 ft2 = 4, 291 in2 (8.50)

55 Each flange has an area of

AF = (0.62)(12.46) = 7.73 ft2 = 1, 113 in2 (8.51)

and the effective depth of the section is d = 12.79 ft, Fig. 8.26.

56 At the crown/hinge the section is rectangular with

Acr = (1.05)(11.48) = 12.05 ft2 = 1, 735 in2 (8.52)


Draft8.5 Salginatobel Bridge (Maillart) 8–19

20 ft 20 ft 20 ft 20 ft 20 ft87.5 ft87.5 ft

295 ft

20 ft

42.6

ft

Figure 8.23: Salginatobel Bridge; Dimensions, (Billington and Mark 1983)

ACTUAL ARCH WITH CENTROID (DOTTED LINE)

IDEALIZATION(ONE DEMENSIONAL)

295 ft

42.6

ft

Figure 8.24: Salginatobel Bridge; Idealization, (Billington and Mark 1983)

CORK PADS

HINGE

ACTUAL SPRINGING HINGE IDEALIZATION

CONCRETE

CONCRETE HINGE

ACTUAL CROWN HINGE IDEALIZATION

CORK PAD

HARD WOOD

Figure 8.25: Salginatobel Bridge; Hinges, (Billington and Mark 1983)



12.46 ft

0.62 ft

0.62 ft

0.59 ft

13.4

1 ft

d=12

.79

ft12

.17

ft

ACTUAL ARCH

SECTIONS295 ft

42.6

ft

Figure 8.26: Salginatobel Bridge; Sections, (Billington and Mark 1983)

8.5.2 Loads

57 The dead load WD is assumed to be linearly distributed (even though it is greater where the arch isdeeper, and the vertical members longer) and is equal to 1,680 kips, Fig. 8.27.

wD =WD

L=

(1, 680) k

(295) ft= 5.7 k/ft (8.53)

L = 295 ft

D

D

W = 1680 k

w = 5.7 k/ft

Figure 8.27: Salginatobel Bridge; Dead Load, (Billington and Mark 1983)

58 For the sake of simplicity we will neglect the snow load (which is actually negligible compared to thedead load).

59 The live load is caused by traffic, and we consider the case in which two trucks, each weighing 55kips, are placed at the quarter-point, Fig. 8.28. This placement of the load actually corresponds to one



of the most critical loading arrangement. The total vertical load is shown in Fig. 8.29

8.5.3 Reactions

60 Reactions are easily determined from equilibrium, Fig. 8.33

VD =1, 680

2= 840 k (8.54-a)

VL =1102

= 55 k (8.54-b)

(+ �✁✛) ΣMc = 0 (8.54-c)

(840)(147.5)− (840)(73.75)−HD(42.6) = 0 (8.54-d)

⇒ HD = 1, 455 k (8.54-e)

(55)(147.5)− (55)(73.75)−HL(42.6) = 0 (8.54-f)

⇒ HL = 95 k (8.54-g)

RD =√

(840)2 + (1, 455)2 = 1, 680 k (8.54-h)

RL =√

(55)2 + (95)2 = 110 k (8.54-i)


61 The shear diagrams for the dead, live and combined load is shown in Fig. 8.31.

62 At the quarter point the axial force can be expressed as:

N = H cosα+ V sinα (8.55)

wheretanα =

2dL

(8.56)

and α = 16.1o at this location. The horizontal force for the dead and live loads was determined previouslyas 1, 455 and 95 kips respectively, and the vertical forces are obtained from the shear diagram, thus

N qrD = (−1, 455) cos16.1o + (−420) sin 16.1o = −1, 514 k (8.57-a)

N qrL = (−95) cos 16.1o + (−55) sin 16.1o = −106 k (8.57-b)

and at the crown where there is no vertical force (and α = 0)

N crD = (−1, 455) cos0o + (−420) sin 0o = −1, 455 k (8.58-a)

N crL = (−95) cos 0o + (−55) sin 0o = −95 k (8.58-b)

63 The uniform dead load will not produce a moment on the parabolic arch.

64 The (point) live load will create a moment which can be decomposed into two parts,

1. Vertical load will cause a trapezoidal moment diagram, and the max moment is

MVL =

P

2L

4=

11o2

2954

= 4, 050 k.ft (8.59)

2. The second is caused by the horizontal reaction, and the resulting moment is MHL = Hd(x), since

d varies parabolically, and H is constant, that second moment is parabolic with a peak value equalto

MVL = Hdmax = (95)(32.6) = −4, 050 k.ft (8.60)



P =

55

kP

= 5

5 k

AR

CH

AB

UT

ME

NT

RO

AD

WA

Y

PLA

N

42.5 ft

295

ft

Figure 8.28: Salginatobel Bridge; Truck Load, (Billington and Mark 1983)



V = 55 kV = 55 kA,L B,LLIVE LOAD

P = 55 k P = 55 k

AB 42

.6 f

t

295 ft

V = 840 k V = 840 kA,D B,D

DQ = 1680 k

DEAD LOAD

AB

Figure 8.29: Salginatobel Bridge; Total Vertical Load, (Billington and Mark 1983)

A

V

H

l/2=147.5 ft

l/4=73.75 ft

d=42.6 ft

CP

Figure 8.30: Salginatobel Bridge; Reactions, (Billington and Mark 1983)

295 ft

295 ft

295 ft

0L/4 L/2

3L/4

840 k

420 k

420 k

840 k

295 ftx

x

+ 420 k

+ 895 k

+ 475 k

- 475 k

- 895 k

- 420 k

0 L

+

=

L/255 k

-55 k

L0

SHE

AR

FO

RC

E

SHE

AR

FO

RC

E

SHE

AR

FO

RC

E

Figure 8.31: Salganitobel Bridge; Shear Diagrams, (Billington and Mark 1983)



at the quarter point

MVL = Hd1/4 = (95)

3(32.6)4

= −3, 040 k.ft (8.61)

65 The overall bending moment diagram from the live loads is determined by simply adding those twocomponents, Fig. 8.32.

MO

ME

NT

BE

ND

ING

295 ft

BE

ND

ING

295 ft

MO

ME

NT

BE

ND

ING

L/4L/4

P P

xL/4 L/2 3L/4 L0

MO

ME

NT

+

=+ PL/4 = 4,050 k-ft

-4,050 k-ft

-3,040 k-ft

4,050 ft.k-3,040 ft.k = 1,010 ft.k

Figure 8.32: Salginatobel Bridge; Live Load Moment Diagram, (Billington and Mark 1983)

66 We observe that the actual shape of the arch follows this bending moment diagram for one of themost critical live load case.

67 The maximum moment at midspan is

MmaxL = 4, 050− 3, 040 = 1, 010 k.ft (8.62)

which would produce internal forces in the upper and lower flanges equal to:

Fint = ±MmaxL

d=

(1, 010) k.ft

(12.8) ft= ±79 k (8.63)


68 The axial stresses at the springlines were determined to be −1, 680 and −110 kips for the dead andlive loads respectively.

69 At the support the area of concrete is Ac = 2, 240 in2, thus the axial stresses are

σDsp =−(1, 680) k

(2, 240) in21, 000 = −750 psi (8.64-a)

σLsp =−(110) k

(2, 240) in21, 000 = −49 psi (8.64-b)

σsspTotal = −750− 49 = −799 psi (8.64-c)

70 At the crown, we repeat the same calculations, where the axial force is equal to the horizontalcomponent of the reactions

σDcr =−(1, 455) k

(1, 735) in21, 000 = −839 psi (8.65-a)



σLcr =−(95) k

(1, 735) in21, 000 = −55 psi (8.65-b)

σscrTotal = −839− 55 = −894 psi (8.65-c)

71 The stresses at the quarter point are determined next. Note that we must include the effect of bothaxial and flexural stresses

σtopqr =−(1, 514) k

(4, 291) in21, 000︸︷︷︸

DeadLoad

+−(106) k

(4, 291) in21, 000︸︷︷︸

LiveLoad︸︷︷︸AxialStresses

− (79) k

(1, 113) in2︸︷︷︸Flexural

(8.66-a)

= −353− 25− 71 −449 psi (8.66-b)

σbotqr =−(1, 514) k

(4, 291) in21, 000︸︷︷︸

DeadLoad

+−(106) k

(4, 291) in21, 000︸︷︷︸

LiveLoad︸︷︷︸AxialStresses

(79) k

(1, 113) in2︸︷︷︸Flexural

(8.66-c)

= −353− 25 + 71 −307 psi (8.66-d)

8.5.6 Structural Behavior of Deck-Stiffened Arches

From (Billington 1979)INCOMPLETE

72 The issue of unsymmetrical live load on a stiffened or unstiffened arch was also addressed by Maillart.As discussed in (Billington 1979) and illustrated by Fig. 8.33

wL

wL

wL a2

wL a2

wL

wL

wL a

wL

Unstiffened Arch Stiffened Arch

∼∆/10∆

a

0

Figure 8.33: Structural Behavior of Stiffened Arches, (Billington 1979)


Draft

Chapter 9

STEEL TENSION MEMBERS

1 This will be your first exposure to structural design. Contrarily to structural analysis where you haveone and only one exact solution, there are typically many acceptable and correct alternative designs.

2 Another important difference, is that in analysis you have so far focused on the application of oneequation (such as the flexural stress in a beam) or of a single very well defined sequential process (suchas truss analysis). In design, (and very much so in the design of tension steel members), you will haveto satisfy more than one equation.

3 In almost all cases, design is governed by a code. In this course we adopt the LRFD method of theAISC code. Hence, a design would have to simultaneously satisfy all the relevant code provisions.

4 In analysis you did not have to concern yourself with details on how the structure is put together,evidently in design you would have to take this into account. To complicate matters, you would have touse some form of common sense to make certain decisions which govern the design process.

5 Finally, design is loosely defined as a process in which you either select new members, or you verifywhether an existing design is satisfactory.

9.1 Introduction

6 Tension members are encountered mostly in truss structures (bridges, roof trusses, transmission towers,wind bracing of multistoried buildings).

7 We will separately cover the design of compression members due to the possibility of buckling.

8 Structural elements are connected together to form a structure. Connections fall into one of twocategories With fasteners, bolts, rivets; or without fasteners, welds. Fasteners require drilling or punchinga hole.

9 It can be shown that a circular hole in a uniformly stressed plate under σ, will have a stress concen-tration factor equal to 3, Fig. 9.1. That is around the hole, σmax = 3σ1.

9.2 Geometric Considerations

10 Before we consider the design of tension members, due consideration must be given to

1. The definition of the cross sectional area which will be used to dimension our design

2. Stiffness of the member1If the circle becomes an ellipse, than the stress concentration factor becomes greater than three; At the limit if the

ellipse collapses into a crack, then we would have theoretically an infinite stress. This is why one should always avoid sharpcorners in a structure.

Draft9–2 STEEL TENSION MEMBERS

Figure 9.1: Stress Concentration Around Circular Hole

9.2.1 Areas

11 The presence of a hole, and the resulting stress concentration must be accounted for in design. Thisis accomplished by introducing the concept of effective areas.

12 The effective area Ae is defined in terms of the net area An when the load is transmitted by boltsor rivets, and gross area Ag for welded members:

Ae = UAn Bolts, RivetsAe = UAg Welds

(9.1)

where U is a reduction factor, and Ag is the gross area of the member.

9.2.1.1 Net Area, An

13 The net area must be defined whenever we have holes due to bolts or rivets.

14 Holes are usually punched 1/16 in. larger than the diameter of the bolt or rivet.

15 When holes are punched, a small zone (∆R ≈ 132 in) around it is damaged, thus the total width to

be deducted in determining the net area is ∆D = 116 in.

16 Thus with respect to the fastener of diameter D we punched a hole of diameter D + 116 , and an

additional 116 was damaged by the process, thus for net area calculation we must subtract 1

8 from thefastener diameter, Fig. 9.2.

17 Accordingly, for a plate of width w and thickness t, and with a single hole which accomodates a boltof diameter D, the net area would be

An = wt−(D +

18

)t (9.2)

18 Whenever there is a need for more than two rivets or bolts, the holes are not aligned but ratherstaggered. Thus there is more than one potential failure line which can be used to determine the netarea.

19 For staggered holes, we define the net are in terms of, Fig. 9.3,

pitch s, longitudinal center to center spacing between two consecutive holes.

gage g, transverse center to center spacing between two adjacent holes.


Draft9.2 Geometric Considerations 9–3

��

��

��

��

Effective hole size

Bolt size

Hole sizeD

D +

D +

161

81

Figure 9.2: Hole Sizes

Figure 9.3: Effect of Staggered Holes on Net Area



The net area is determined as above, however we subtract

Length Correction =s2

4g(9.3)

An = Ag − Σ# of holes1

(D +

18

)t+ Σ# of correct.

1

s2

4gt (9.4)

In this example the two possible failure lines have the following net lengths:

A-B =Length of A-B −(width of hole + 116 in.)

A-C =Length of A-B −2(width of hole + 116 in.) + s2

4g

The minimum net area will obviously be equal to the net length multiplied by the plate thickness.

20 When holes are staggered on two legs of an angle, the gage length g is obtained by using a lengthbetween the centers of the holes measured along the centerline of the angle thickness, i.e., the distance

A-B in Fig. 9.4. Thus the gage distance is g = ga − t2 + gb − t2 or

Figure 9.4: Gage Distances for an Angle

g = ga + gb − t (9.5)

21 It should be mentioned that every rolled angle has a standard value for the location of holes (i.e.gage distances ga and gb) depending on the length of the leg, Table 9.1.

Example 9-1: Net Area

Determine the minimum net area of the plate shown below assuming 1516 in.-diameter holes are located

as shown.



Leg 8 7 6 5 4 312 3 2

12 2 1

34 1

12 1

38 1

14 1

g 412 4 3

12 3 2

12 2 1

34 1

38 1

18 1 7

878

34

58

g1 3 212 2

14 2

g2 3 3 212 1

34

Table 9.1: Usual Gage Lengths g, g1, g2

Solution:We consider various paths:

A−D :[12− 2

(1516 + 1

16

)](0.25) = 2.50 in2 (9.6-a)

A−B −D :[12− 3

(1516 + 1

16

)+ (2.125)2

4(2.5) + (2.125)2

4(4)

](0.25) = 2.43 in2 (9.6-b)

A−B − C :[12− 3

(1516 + 1

16

)+ (2.125)2

4(2.5) + (1.875)2

4(4)

](0.25) = 2.4 in2 (9.6-c)

Thus path A-B-C controls.

Example 9-2: Net Area, Angle

Determine the net area An for the angle shown below if 1516 in. diameter holes are used.

Solution:For the net area calculation, the angle may be visualized as being flattened into a plate as shown below.



and

An = Ag −Dt+ s2

4gt (9.7)

Where D is the width to be deducted for the hole, We consider the following paths:

A− C : 4.75− 2(1516 + 1

16

)0.5 = 3.75 in2 (9.8-a)

A− B − C : 4.75− 3(1516 + 1

16

)0.5 +

[(3)2

4(2.5) + (3)2

4(4.25)

]0.5 = 3.96 in2 (9.8-b)

9.2.1.2 Effective Net Area, Ae

22 Often tension members are connected at the end by connection to some but not all of the cross sectionalelements. For instance, an angle section may be connected through one leg only. Thus, away from theconnection we used the gross or net area to resist the full tensile load, but right at the connection, onlyone leg is connected and the other is for all practical purposes unstressed.

23 To account for this, the AISC code stipulates that the net area must be reduced by U , Table 9.2where Ae = UAn or Ae = UAg.

Example 9-3: Reduction Factor

Determine the reduction factor U to be used in computing the effective net area of a W14x82 sectionconnected by plates at its two flanges. There are three bolts along each connection line.Solution:Since only two (the flanges) of the three elements (the third one being the web) are connected, U < 1.We first check for the profile ratio2 bf

d .

bfd

=10.1314.31

= 0.71 > 0.67√

(9.9)

Thus from Table 9.2, U = 0.9

9.2.2 Stiffness

24 Tension members which are too long may be

1. Too flexible during erection of the structure (excessive and possibly permanent deformations whencarried due to their own weight).

2Section properties are taken from Sect. 3.6.



Type of members Minimum Number Special Aeof Fasteners/line Requirements

Fastened Rolled SectionsMembers having all cross sectional elements con-

nected to transmit tensile force

1 (or welds) An

W, M, or S shapes with flange widths not less

than 2/3 the depth, and structural tees cut from

these shapes, provided the connection is to the

flanges. Bolted or riveted connections shall have

no fewer than three fasteners per line in the di-

rection of stress

3 (or welds) bd ≥ 0.67 0.9An

W, M, or S not meeting above conditions, struc-

tural tees cut from these shapes and all other

shapes, including built-up cross-sections. Bolted

or riveted connections shall have no fewer than

three fasteners per line in the direction of stress

3 (or welds) 0.85An

Structural tees cut from W, M, or S connected at

flanges only

3 (or welds) bd ≥ 0.67 0.9An

All members with bolted or riveted connections

having only two fasteners per line in the direction

of stress

2 0.75An

Welded PlatesWelds l > 2w AgWelds 2w > l > 1.5w 0.87AgWelds l/w.1 0.75Ag

b Flange width; d section depth; l Weld length; w Plate width;

Table 9.2: Effective Net Area Ae for Bolted and Welded Connections



2. Subjected to excessive lateral deflections and vibrations once assembled.

25 To limit this, the AISC code imposes a constraint on the element stiffness.

26 Recall that stiffness is defined as K = P∆ , and for flexural member it will be shown later that

Kf = Mθ = 4EI

L where I is the moment of inertia.

27 Since for steel E, the elastic modulus is constant, and given the definition of the radius of gyration

r =√

IA , the AISC code (LRFD-B7) specifies that the slenderness ratio L

rminshould not exceed 300

L

rmin< 300 (9.10)

This limitation does not apply to rods in tension.

28 In applying this requirement, the higher slenderness ratio based on the two principal axes must beused, i.e. L

rxand L

ry.

9.3 LRFD Design of Tension Members

29 Recall the general form of the structural safety requirement under LRFD (Eq. 7.9):

ΦRn ≥ ΣαiQi (9.11)

For tension members, this equation becomes

ΦtTn ≥ Tu (9.12)

where Φt = resistance factor relating to tensile strengthTn = nominal strength of a tension memberTu = factored load on a tension member

30 From Table 7.3, Φt is 0.75 and 0.9 for fracture and yielding failure respectively.

9.3.1 Tension Failure

31 The design strength ΦtTn is the smaller of that based on, Fig. 9.5

1. Yielding in the gross section: We can not allow yielding of the gross section, because this willresult in unacceptable elongation of the entire member.

ΦtTn = ΦtFyAg = 0.90FyAg (9.13)

or

2. Fracture in the net section: Yielding is locally allowed, because Ae is applicable only on a smallportion of the element. Local excessive elongation is allowed, however fracture must be prevented.This mode of failure usually occurs if there is insufficient distance behind the pin.

ΦtTn = ΦtFuAe = 0.75FuAe (9.14)


Draft9.3 LRFD Design of Tension Members 9–9

oo

oo

o

ooo

o

o

oo

oo

oooo

ooo o

o oo

0.75F A 0.9F A 0.75F A u e u ey g

Figure 9.5: Net and Gross Areas

9.3.2 Block Shear Failure

32 For bolted connections, tearing failure may occur and control the strength of the tension member.

33 For instance, with reference to Fig. 9.6 the angle tension member attached to the gusset plate3 may

��

��

��

��

��

��

��

��

Figure 9.6: Tearing Failure Limit State

have a tearing failure along a-b-c. In this case the shear strength along a-b plus the tensile strengthalong bc contribute to resist this particular failure mode.

34 This mode of failure is uncommon in tension members; However this failure mode commonly controlsthe design of bolted end connections to the thin webs of beams.

35 LRFD-J4 indicates that the following two equations must be considered to account for either one ofthe two combinations

Shear yielding-tension fracture: fracture on the net area in tension followed by yielding on the gross3A gusset plate is a plate at the intersection of members to which they are connected.



section in shear

ΦtTn = 0.75(0.6FyAvg + FuAnt) (9.15)

Shear fracture -tension yielding: Fracture on the net section in shear followed by yielding on thegross section in tension.

ΦtTn = 0.75(0.6FuAns + FyAtg) (9.16)

where Avg = gross area subjected to shear yieldingAtg = gross area subjected to tensile yieldingAns = net (gross - area of holes) area subjected to shear fractureAnt = net (gross - area of holes) area subjected to tensile fracture

and the largest of the two will control.

36 Note that a conservative simplification of the above two equations is given below

ΦtTn = 0.75(0.6Fu)Ans (9.17)

In this equation Ans is the area along the entire failure path (including both tension and shear failure),whereas in Eq. 9.15 and 9.16 Ans is the net area where actual shear failure occurs (not including tensionfailure).

37 Note that in the preceding equation, we assumed the shear yield stress τy = 0.6Fy, and the sheartensile strength τu = 0.6Fu.

9.3.3 Summary

38 The design procedure for tension member can be summarized as follows

1. Determine the net area An from Section 9.2.1.1; if there are staggered holes, use Eq. 9.3.

2. Determine the effective area Ae from Section 9.2.1.2, and U from Table 9.2.

3. The limit load will be the lowest of the following

Tension Failure: by taking the lowest of the following

(a) Yielding (σyld) in the gross section, Eq. 9.13(b) Fracture (σu) in the effective section, Eq. 9.14

Tearing: By checking either one of the following

(a) Eq. 9.17(b) Eq. 9.15 and 9.16, take the largest value.

4. Check for slenderness, Eq. 9.10

Example 9-4: Load Capacity of Angle

Determine the service load capacity in tension for an L6x4x12 of A572 grade 50 steel connected with

78 in. diameter bolts as shown below. Assume a live load to dead load ratio of 3.0. Since there are manyfasteners, there is no need to check for shear failure.



Solution:

1. The net area is given by

(a) Section 1-1:

An = Ag − 1 hole = 4.75−(

78

+18

)0.50 = 4.25 in2 (9.18)

(b) Section 1-2:

An = Ag − 2 holes +(s2

4g

)t (9.19-a)

= 4.75− 2(

78

+18

)0.50 +

(2)2

4(2.5)(0.5) = 3.95 in2 (9.19-b)

2. Since not both legs are connected, From Table 9.2 U = 0.85 and

Ae = UAn = (0.85)(3.95) = 3.36 in2 (9.20)

3. Checking yielding failure of the gross area (Eq. 9.13)

ΦtTn = ΦtFyAg (9.21-a)= 0.90(50)(4.75) = 214 k (9.21-b)

4. Checking for fracture in the net section, Eq. 9.14

ΦtTn = ΦtFuAe (9.22-a)

= 0.75(65)(3.36) = 163.8 k (9.22-b)

5. Thus the controlling ΦtTn is 163.8 Kips

6. Next we must determine the factored load, From Section 7.3.3 we consider ΦTn = 1.2D + 1.6L,since L = 3D, this reduces to ΦTn = 1.2D + 1.6(3D) = 6.0D or

ΦTn = 163.8 = 6.0D (9.23-a)

D = 27.3 k (9.23-b)

L = (3)(27.3) = 81.9 k (9.23-c)



Example 9-5: Shear Rupture

Investigate the shear rupture failure mode on the angle L4x4x14 attached with three 7

8 in. diameterbolts to a 3

8 in. gusset plate. The material is A36 steel.

Solution:


ΦtTn = ΦtFyAg (9.24-a)= 0.90(36)(1.94) (9.24-b)= 62.9 k (9.24-c)

2. Checking for fracture in the net section, Eq. (9.14)

ΦtTn = ΦtFuAe = ΦtFuUAn (9.25-a)

= 0.75(58)(0.85)[1.94− (78

+18)(0.25)] (9.25-b)

= 62.5 k (9.25-c)

3. Thus the controlling ΦtTn is 62.5 Kips

4. The block shear failure along path a-b-c investigated according to LRFD-J4 using Eq. 9.17 gives

Ans = (length a-b-c less three holes)thickness (9.26-a)

=[7.5 + 1.5− (3)(

78

+18)]0.25 (9.26-b)

= 1.5 in2 (9.26-c)ΦtTn = Φ(0.6Fu)Ans (9.26-d)

= 0.75(0.6)(58)(1.5) (9.26-e)

= 39.15 k (9.26-f)

Thus, block shear gives a lower design strength than yielding or fracture.

5. Had we instead used the larger of Eq. 9.15 and 9.16 we would have obtained



(a) Shear yielding-tension fracture:

ΦtTn = Φt(0.6FyAvg + FuAnt) (9.27-a)

= 0.75{

(0.6)(36) [(2× 3) + 1.5] (0.25) + (58)[1.5− 1

2(78

+18)](0.25)

}(9.27-b)

= 41.25 k (9.27-c)

(b) Shear fracture -tension yielding:

ΦtTn = Φt(0.6FuAns + FyAtg) (9.28-a)

= 0.75{

(0.6)(58)[7.5− (

78

+18)(2.5)

](0.25) + (36)(1.5)(0.25)

}(9.28-b)

= 42.75 k (9.28-c)

6. Hence, when the more correct equations are used, the strength is controlled by block shear; ΦtTn =43.8 k which is less than the fracture strength of 62.5 k.

7. This general result is expected when minimum number of connectors are used along with thinelements such as 1

4 in. or less.

Example 9-6: Shear Rupture

Determine the service load that can be carried by this C15x33.9 section (Ag = 9.96 in2), assumingthat the load is 20% dead load, and 80% live load. A36 steel with 3/4 in. diameter bolts.

1.5" 3" 3"

3"

3"

3"

15"

C15X33.9

t =0.4"w

Solution:


ΦtTn = ΦtFyAg (9.29-a)= 0.90(36)(9.96) = 322 k (9.29-b)

(9.29-c)

2. Checking for fracture in the net section, Eq. (9.14)

An = = 9.96− 4(34

+18)(0.4) = 9.96− 4(0.875)(0.4) = 8.56 in2 (9.30-a)

Ae = UAn = 0.85(8.56) = 7.276 in2 (9.30-b)ΦtTn = ΦtFuAe = ΦtFuUAn (9.30-c)

= 0.75(58)(0.85)(7.276)] = 269. k (9.30-d)



3. Thus the controlling ΦtTn is 269 Kips

4. The block shear failure according to LRFD-J4 using Eq. 9.17 gives

Ans = [2(7.5) + 9− 8(0.875)]0.4 = 6.80 in2 (9.31-a)ΦtTn = Φ(0.6Fu)Ans (9.31-b)

= 0.75(0.6)(58)(6.8) (9.31-c)

= 177.5 k (9.31-d)

Thus, block shear gives a lower design strength than yielding or fracture.

5. Had we instead used the larger of Eq. 9.15 and 9.16 we would have obtained

(a) Shear yielding-tension fracture:

Avg = 2(7.5)(0.4) = 6.00 in2 (9.32-a)Ant = [9− 3(0.875)](0.4) = 2.55 in2 (9.32-b)

ΦtTn = Φt(0.6FyAvg + FuAnt) (9.32-c)= 0.75[(0.6)(36)(6.0) + (58)(2.55)] = 208 in2 (9.32-d)

(9.32-e)

(b) Shear fracture -tension yielding:

Ans = 2[7.5− (2.5)(0.875)](0.4) = 4.25 in2 (9.33-a)Atg = (9)(0.4) = 3.60 in2 (9.33-b)

ΦtTn = Φt(0.6FuAns + FyAtg) (9.33-c)

= 0.75[(0.6)(58)(4.25)+ (36)(3.6)] = 208 k ✛ (9.33-d)

6. Thus the design tension T − d is 208 kips

7. Finally we determine the service loads

DL = 0.2P (9.34-a)LL = 0.8P (9.34-b)

1.4DL = (1.4)(0.2)P = 0.28P (9.34-c)1.2DL+ 1.6LL = (1.2)(0.2P ) + (1.6)(0.8P ) = 1.52P ✛ (9.34-d)

ΦtTn = 208 k = 1.52P ⇒ P = 137 k (9.34-e)

DL = (0.2)(137) = 27 k (9.34-f)

LL = (0.8)(137) = 109 k (9.34-g)

Note that P is the sum of the service loads (DL +LL).

Example 9-7: Complete Analysis/Design of a Truss

Design the tension members of a welded roof truss for the industrial building made of A36 steel.Trusses are space 20 ft center to center.Roof loads: 1) Built-up roofing: 5.5 psf; 2) Decking: 2.5 psf; 3) Purlins: 3.0 psf; 4) Truss (estimate)

3.0 psf.Snow load 40 psf.Trolley load on bottom chord: 6.4 k at L2 and L′

2; and 12.3 k at L5.



Solution:

Loading: First we must determine the joint load in order to analyze the truss.

1. The total roof load is5.5 + 2.5 + 3.0 + 3.0 = 14.0 psf (9.35)

Top chord loads:

D = (14.0) psf(20) ft = 280 lbs/ft (9.36-a)S = (40.0) psf(20) ft = 800 lbs/ft (9.36-b)

2. Next we must translate this vertical uniform load into joint loads on the truss

U0, U′0 : Snow Load: 1

2 (6.42)(800)1

1,000 = 2.57 k (9.37-a)

Roof Load: 12 (6.42)(280)

11,000 = 0.9 k (9.37-b)

U1, U′1 : Snow Load: 1

2 (6.42 + 6.00)(800) 11,000 = 4.97 k (9.37-c)

Roof Load: 12 (6.42 + 6.00)(280) 1

1,000 = 1.74 k (9.37-d)

U2, U′2, U3, U

′3, U4, U

′4, U4 : Snow Load: (6.00)(800) 1

1,000 = 4.80 k (9.37-e)

Roof Load: (6.00)(280) 11,000 = 1.68 k (9.37-f)

Analysis: With the load defined, we are now ready to perform a structural analysis of the truss:

1. We have three types of loads (Dead, Snow, and Live), and according to the AISC code (Sect.7.3.3 we must consider the following three load conditions:

(a) 1.4 D(b) 1.2D+1.6L+0.5S(c) 1.2D+1.6S+0.5L

2. We have 22 joints, 3 reactions and 41 members; 22(2) = 41 + 3√

thus the truss is staticallydeterminate.

3. In this case, the structure is symmetrical. That is there is no horizontal displacements at jointsL5 and U5, however those joints undergo a vertical displacement. Hence, we can analyze theleft half of the truss, and assume that we do have roller supports along the y axis at joints



L5 and U5. This substructure will have 3 reactions, 21 elements, and 12 joints. Thus, it isstatically determinate.Note that at joint U5 we will apply only one half of the load due to symmetry.

4. Hence, we shall analyze the structure for each load (D, S, and L), and then determine thevarious load combinations. This is best achieved by the Matrix method described in Sect.5.2.2.1.

5. We renumber the truss as shown below.

6. Let us determine the direction cosines for each member4.

Member joint ∆x ∆y L α β Joint ∆x ∆y α β1 1 6.42 0. 6.42 1.0 0. 3 -6.42 0. -1.0 0.2 1 0 4.50 4.50 0. 1.0 2 0. -4.50 0. -1.03 2 6.42 -4.50 7.84 .82 -.57 3 -6.42 4.50 -.82 .574 2 6.42 0.54 6.44 1.0 .08 4 -6.42 -0.54 -1.0 -.085 3 0. 5.04 5.04 0. 1.0 4 0. -5.04 0. -1.06 3 6.0 0. 6.0 1.0 0. 5 -6.0 0. -1.0 0.7 4 6.0 -5.04 7.84 .77 -.64 5 -6.0 5.04 -.77 .648 4 6.0 .501 6.02 1.0 .08 6 -6.0 -.501 -1.0 -.089 5 0. 5.54 5.54 0. 1.0 6 0. -5.54 0. -1.010 5 6.0 0. 6.0 1.0 0. 7 -6.0 0. -1.0 0.11 6 6.0 -5.54 8.17 .73 -.68 7 -6.0 5.54 -.73 .6812 6 6.0 .501 6.02 1.0 .08 8 -6.0 -.501 -1.0 -.0813 7 0. 6.04 6.04 0. 1.0 8 0. -6.04 0. -114 7 6.0 0. 6.0 1.0 0. 9 -6.0 0. -1.0 0.15 8 6.0 -6.04 8.51 .71 -.71 9 -6.0 6.04 -.71 .7116 8 6.0 .501 6.02 1.0 .08 10 -6.0 -.501 -1.0 -.0817 9 0. 6.54 6.54 0. 1.0 10 0. -6.54 0. -1.018 9 6.0 0. 6.0 1.0 0. 11 -6.0 0. -1.0 0.19 10 6.0 -6.54 8.88 .68 -.74 11 -6.0 6.54 -.68 .7420 10 6.0 .501 6.02 1.0 .08 12 -6.0 -.501 -1.0 -.0821 11 0. 7.04 7.04 0. 1.0 12 0. -7.04 0. -1.0

7. Having determined the directions cosines,and assuming all members under tension, the statics[B] matrix can be assembled5.

[B] = [BL|BR]4Note that the upper chord have an elevation of y = 4.5 + .08356x with respect to L0.5Note that in assembling the matrix, each column corresponds to an unknown internal element force, or an external

reaction. Each row corresponds to an equation of equilibrium. To facilitate the construction of the matrix, equations ofequilibrium at each joint are lumped into a “row” containing two lines. The first line corresponds to ΣFx, and the secondone to ΣFy. Thus, the first line contains αu

j , and the second βuj where the subscript corresponds to the joint (or “row”

number). The superscript corresponds to the column number or unknown quantity (member force or reaction). αuj and

βuj are taken from the table above.



or

[BL] =

# 1 ΣFxΣFy

# 2 ΣFxΣFy

# 3 ΣFxΣFy

# 4 ΣFxΣFy

# 5 ΣFxΣFy

# 6 ΣFxΣFy

# 7 ΣFxΣFy

# 8 ΣFxΣFy

# 9 ΣFxΣFy

# 10 ΣFxΣFy

# 11 ΣFxΣFy

# 12 ΣFxΣFy

1. 0.0. 1.

0. .82 1.−1. −.57 .08

−1. −.82 0. 1.0. .57 1. 0.

−1. 0. −.77 1.−.08 −1. .64 .08

−1. .77 0. 1.0. −.64 1. 0.

−1. 0. .73 1.−.08 −1. −.68 .08

−1. −.730. .68

−1.−.08

[BR] =

# 1 ΣFxΣFy

# 2 ΣFxΣFy

# 3 ΣFxΣFy

# 4 ΣFxΣFy

# 5 ΣFxΣFy

# 6 ΣFxΣFy

# 7 ΣFxΣFy

# 8 ΣFxΣFy

# 9 ΣFxΣFy

# 10 ΣFxΣFy

# 11 ΣFxΣFy

# 12 ΣFxΣFy

1.

0. 1.1. 0.0. .71 1.

−1. −.71 .08−1. −.71 0. 1.0. .71 1. 0.

−1. 0. .68 1.0−.08 −1. −.74 .08

−1. −.68 0. 1.0. .74 1.

−1. 0. 1.−.08 −1.

F1F2F3F4F5F6F7F8F9F10F11F12F13F14F15F16F17F18F19F20F21R1yR11xR12x

8. We will have three vectors of joint loads

P1xP1yP2xP2yP3xP3yP4xP4yP5xP5yP6xP6yP7xP7yP8xP8yP9xP9yP10xP10yP12xP12y

= α1

0.0.0.

−0.9 k0.0.0.

−1.74 k0.0.0.

−1.68 k0.0.0.

−1.68 k0.0.0.

−1.68 k0.0.0.

−0.84 k

︸︷︷︸

Roof

+α2

0.0.0.

−2.57 k0.0.0.

−4.97 k0.0.0.

−4.8 k0.0.0.

−4.8 k0.0.0.

−4.8 k0.0.0.

−2.4 k

︸︷︷︸

Snow

+α3

0.0.0.0.0.0.0.0.0.

−6.40.0.0.0.0.0.0.0.0.0.0.

−6.150.0.

︸︷︷︸

Crane (Live Load)

9. There are many tools available to solve this problem. We will use Mathematica. Below is alisting of our input file.(* define the statics matrix*)b={{ 1., 0. , 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.},{ 0., 1. , 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,0., 0., 0., 0., 0., 0., 0., 0., 0., 1. , 0., 0.},{ 0., 0. , .82, 1. , 0., 0., 0., 0., 0., 0., 0., 0.,0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.},{ 0., -1.,-.57, .08, 0., 0., 0., 0., 0., 0., 0., 0.,0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.},{-1., 0.,-.82, 0., 0. , 1. , 0., 0., 0., 0., 0., 0.,0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.},{ 0., 0., .57, 0., 1. , 0. , 0., 0., 0., 0., 0., 0.,0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.},



{ 0., 0., 0., -1., 0. , 0.,-.77, 1. , 0., 0., 0., 0.,0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.},{ 0., 0., 0.,-.08, -1., 0., .64, .08, 0., 0., 0., 0.,0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.},{ 0., 0., 0., 0., 0., -1., .77, 0., 0. , 1. , 0., 0.,0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.},{ 0., 0., 0., 0., 0., 0.,-.64, 0., 1. , 0. , 0., 0.,0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.},{ 0., 0., 0., 0., 0., 0., 0.,-1. , 0. , 0.,.73 , 1.,0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.},{ 0., 0., 0., 0., 0., 0., 0.,-.08, -1., 0.,-.68, .08,0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.},{ 0., 0., 0., 0., 0., 0., 0., 0., 0., -1.,-.73, 0.,0., 1. , 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.},{ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0. ,.68 , 0.,1., 0. , 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.},{ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., -1.,0., 0., .71, 1., 0., 0., 0., 0., 0., 0., 0., 0.},{ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,-.08,-1., 0.,-.71,.08 , 0., 0., 0., 0., 0., 0., 0., 0.},{ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,0., -1.,-.71, 0., 0. , 1. , 0., 0., 0., 0., 0., 0.},{ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,0., 0. , .71, 0., 1. , 0. , 0., 0., 0., 0., 0., 0.},{ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,0., 0., 0., -1., 0. , 0.,.68 , 1.0, 0., 0., 0., 0.},{ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,0., 0., 0.,-.08, -1., 0.,-.74, .08, 0., 0., 0., 0.},{ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,0., 0., 0., 0., 0., -1.,-.68, 0., 0. , 0., 1. , 0.},{ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,0., 0., 0., 0., 0., 0. , .74, 0., 1. , 0., 0., 0.},{ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,0., 0., 0., 0., 0., 0., 0.,-1. , 0. , 0., 0., 1.},{ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,0., 0., 0., 0., 0., 0., 0.,-.08, -1., 0., 0., 0.}}(* define the joint load vectors (roof, snow, and live)*)r=-{0., 0., 0., -0.9, 0., 0., 0., -1.74, 0., 0., 0., -1.68,0., 0., 0., -1.68, 0., 0., 0., -1.68, 0., 0., 0., -0.84}

s=-{0., 0., 0., -2.57, 0., 0., 0., -4.97, 0., 0., 0., -4.8,0., 0., 0., -4.8 , 0., 0., 0., -4.8 , 0., 0., 0., -2.4}

l=-{0., 0., 0., 0., 0., 0., 0., 0., 0., -6.4, 0.,0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,-6.15, 0., 0.}(* Invert the statics matrix b*)bm1=Inverse[b](* Solve for the roof, snow, and live load forces*)rf=N[bm1.r,2]sf=N[bm1.s,2]lf=N[bm1.l,2]Print[" Dead (roof) Load Forces/Reactions"]Print[rf]Print[" Live (Crane) Load Forces/Reactions"]Print[lf]Print[" Snow Load Forces/Reactions"]Print[sf]

(* Consider various load combinations ber AISC code*)lc1=N[1.4 rf,2]lc2=N[1.2 rf + 1.6 lf +0.5 sf,2]lc3=N[1.2 rf + 0.5 lf + 1.6 sf,2]Print[" Load Combination 1: 1.4 D"]Print[lc1]Print[" Load Combination 2: 1.2 D + 1.6 L +0.5 S"]Print[lc2]Print[" Load Combination 3: 1.2 D +0.5 L +1.6 S"]Print[lc3]Print[" Get the Maximum load of each member"]maxvec=Map[Max,Transpose[{lc1,lc2,lc3}]]Print[" Get the Minimum load of each member"]minvec=Map[Min,Transpose[{lc1,lc2,lc3}]](* Function definitions to prune results*)pos[x_]:=If[x>0,x,0]neg[x_]:=If[x<0,x,0](* Get the vector with Maximum and Minimum forces*)tension=Map[pos,maxvec]compression=Map[neg,minvec]



Print[" Vector of Tensile Forces"]Print[tension]Print[" Vector of Compressive Forces"]Print[compression](* Define a function which will give the cross sectional area*)area[x_]:=If[x>0,x/(0.9*36),"-"](* Det ermine the area for each element which is under tension *)xarea=N[Map[area,tempos],2]Print[" Cross sectional areas for elements under tension"]Print[xarea]

10. The following reactions are obtained

Load R1y R11x R12x

Roof 8.5 19. -19.Snow 24 53. -53.Crane 13. 38 -38

We observe that ΣFx = 0√

, we check for ΣFy = 0 for the roof load (Never ever trust anycomputer analysis unless you check at least equilibrium, and the order of magnitude of thedeflections). The external joint load is 0.9 + 1.74 + 1.68 + 1.68 + 1.68 + 0.84 = 8.52

√.

Design: 1. The above results can now be summarized in the following table

Member Load Load Cases Design LoadRoof Snow Live 1 2 3 Compr. Tens.

1 L0 − L1 0. 0. 0. 0. 0. 0. 0 02 U0 − L0 -8.5 -24. -13. -12. -35. -55. -55. 03 U0 − L1 12. 34. 20. 17. 52. 79. 0 79.4 U0 − U1 -9.8 -28. -16. -14. -43. -65. -65. 05 U1 − L1 -6.8 -20. -11. -9.6 -30. -45. -45. 06 L1 − L2 9.8 28. 16. 14. 43. 65. 0. 65.7 U1 − L2 7.3 21. 16. 10. 38. 50. 0. 50.8 U1 − U2 -15. -44. -29. -22. -72. -100. -100. 09 U2 − L2 -4.6 -13. -3.9 -6.5 -14. -29. -29. 010 L2 − L3 15. 44. 29. 22. 72. 100. 0 100.11 U2 − L3 4. 11. 5.2 5.6 15. 26. 0 26.12 U2 − U3 -18. -52. -32. -26. -83. -120. -120. 013 U3 − L3 -2.7 -7.8 -3.6 -3.8 -10. -18. -18. 014 L3 − L4 18. 52. 32. 26. 83. 120. 0 120.15 U3 − L4 1.4 3.9 4.6 1.9 9.8 10. 0 10.16 U3 − U4 -19. -55. -36. -27. -90. -130. -130. 017 U4 − L4 -0.97 -2.8 -3.3 -1.4 -6.9 -7.3 -7.3 018 L4 − L5 19. 55. 36. 27. 90. 130. 0 130.19 U4 − L5 -0.89 -2.5 4.2 -1.2 5.1 -3.1 -3.1 5.120 U4 − U5 -19. -53. -38. -26. -93. -130. -130. 021 U5 − L5 0.66 1.9 3.1 0.92 6. 5.3 0 6.

2. We now focus on the design of members with tensile forces (Note that member U4 −L5) canbe subjected to both tensile and compressive forces).



Member Des. Load Reqd. Area Section Area L rminL

rminkip in2 in2 in. in.

L0 − L1 - - WT 7x15 4.42 77. 1.49 52L1 − L2 65. 2.01 WT 7x15 4.42 72. 1.49 48L2 − L3 100. 3.09 WT 7x15 4.42 72. 1.49 48L3 − L4 120. 3.70 WT 7x15 4.42 72 1.49 48L4 − L5 130. 4.01 WT 7x15 4.42 72 1.49 48U0 − L1 79. 2.44 2L3x2

12 x 1

4 2.63 94. 0.75 125U1 − L2 50. 1.54 2L2x2x 1

4 1.88 94. 0.61 154U2 − L3 26. 0.80 2L2x2x 3

16 1.43 98. 0.62 158U3 − L4 10. 0.31 2L2x2x 3

16 1.43 102. 0.62 165U4 − L5 5.1 0.16 2L2x2x 3

16 1.43 107 0.62 172U5 − L5 6. 0.19 2L2x2x 3

16 1.43 84 0.62 135

Comments:

1. We designed only the members under tension (Note that U4 − L5) would also have to bedesigned for compression.

2. The same size section was used for the entire lower chord. A smaller section could have beenused for L0 −L2, however the cost saving in steel would probably be offset by the additionallabor cost.

3. Similarly, additional savings would result in using different sections for the inclined members.

4. Again we could have used angles smaller than 2L2x2x 316 , for instance 2L2x2x 1

8 , however it ispreferable to avoid thicknesses smaller than 3

16 in. because of welding.

5. Angles with legs smaller than 2 in. could have been used, however those may not always beavailable, and they may result in a slenderness ratio exceeding 300.

6. The joint details of a truss of this kind are usually worked out by the fabricator. How toreview them and approve them, the Engineer must know how to design them.

7. Design of steel connections will be covered in subsequent courses (note that however we haveseen above how to review some aspects of this design).

9.4 †† Computer Aided Analysis

In more advanced Structural Analysis courses (CVEN4525), you will be exposed to the computer aidedanalysis of structures.

Using such a program (which will be made available later to you), following is the input data file:

*TITLEDesign of Welded truss*CONTROL

12 21 1 0 6 2 0*COORD

1 0.0 0.02 0.0 4.53 6.42 0.04 6.42 5.045 12.42 0.06 12.42 5.547 18.42 0.08 18.42 6.049 24.42 0.0

10 24.42 6.5411 30.42 0.012 30.42 7.04

*BOUND1 0 1

11 1 012 1 0

*ELEM1 1 3 12 1 2 1


Draft9.4 †† Computer Aided Analysis 9–21

3 2 3 14 2 4 15 3 4 16 3 5 17 4 5 18 4 6 19 5 6 1

10 5 7 111 6 7 112 6 8 113 7 8 114 7 9 115 8 9 116 8 10 117 9 10 118 9 11 119 10 11 120 10 12 121 11 12 1

*PROP1 30000. 1.

*LOAD3 3

Roof Load6 0 02 0. -0.94 0. -1.746 0. -1.688 0. -1.6810 0. -1.6812 0. -0.84

Snow Load6 0 02 0. -2.574 0. -4.976 0. -4.88 0. -4.810 0. -4.812 0. -2.4

Crane load2 0 05 0. -6.4

11 0. -6.15*COMBLC 1

11 1.4

LC 231 1.22 0.53 1.6

LC 331 1.22 1.63 0.5

The output of such an analysis would be as follow. Compare those results with the one obtained byinverting the statics matrix.

*********************************** LOAD CASE: 5

I S D S

DISPLACEMENTS

INTEGRATED STRUCTURAL DESIGN SYSTEM -------------

*********************************** DEFLECTIONS

DEPT. OF CIVIL ENGINEERING, UNIV. OF COLORADO, BOULDER, CO JOINT -----------------------

Version 0.2 NUMBER DELTA-X DELTA-Y

July 1991 ------ ----------- -----------

1 -6.8194E-02 0.0000E+00

2 4.6579E-02 -6.3711E-03

3 -6.8194E-02 -1.9852E-01

NUMBER NUMBER NUMBER OF NUMBER OF NUMBER OF STRUCTURE 4 5.2197E-02 -2.0454E-01

OF OF PROPERTY ELASTIC LOADING TYPE CODE 5 -5.7975E-02 -3.5366E-01

JOINTS MEMBERS GROUPS SPRINGS CASES NUMBER 6 4.7734E-02 -3.5703E-01

------ ------- --------- --------- --------- --------- 7 -4.0982E-02 -4.6059E-01

12 21 1 0 6 2 8 3.6636E-02 -4.6314E-01

9 -2.1247E-02 -5.2493E-01

10 2.0455E-02 -5.2659E-01

ASSOCIATED D.O.F. NUMBERS AND COORDINATES 11 0.0000E+00 -5.4697E-01

----------------------------------------- 12 0.0000E+00 -5.4536E-01

JOINT DELTA-X DELTA-Y GLOBAL GLOBAL

NUMBER D.O.F.# D.O.F.# X-COOR Y-COOR REACTIONS

------ ------- ------- ------ ------ ---------

1 1 0 0.0 0.0 JOINT FORCE IN X FORCE IN Y

2 2 3 0.0 4.5 NUMBER DIRECTION DIRECTION

3 4 5 6.4 0.0 ------ ---------- ----------

4 6 7 6.4 5.0 1 0.000E+00 4.247E+01

5 8 9 12.4 0.0 11 1.090E+02 0.000E+00

6 10 11 12.4 5.5 12 -1.090E+02 0.000E+00

7 12 13 18.4 0.0

8 14 15 18.4 6.0

9 16 17 24.4 0.0



10 18 19 24.4 6.5 LOAD CASE: 6

11 0 20 30.4 0.0

12 0 21 30.4 7.0 DISPLACEMENTS

-------------

DEFLECTIONS

MEMBER INCIDENCES AND PROPERTY GROUP ID NUMBERS JOINT -----------------------

----------------------------------------------- NUMBER DELTA-X DELTA-Y

MEMBER START END PROPERTY ------ ----------- -----------

NUMBER JOINT JOINT GROUP # 1 -8.2746E-02 0.0000E+00

------ ----- ----- -------- 2 5.5835E-02 -8.3164E-03

1 1 3 1 3 -8.2746E-02 -2.4162E-01

2 1 2 1 4 6.2249E-02 -2.4915E-01

3 2 3 1 5 -6.9944E-02 -4.2656E-01

4 2 4 1 6 5.6902E-02 -4.3183E-01

5 3 4 1 7 -4.9587E-02 -5.5737E-01

6 3 5 1 8 4.3305E-02 -5.6085E-01

7 4 5 1 9 -2.5489E-02 -6.3314E-01

8 4 6 1 10 2.3701E-02 -6.3466E-01

9 5 6 1 11 0.0000E+00 -6.5504E-01

10 5 7 1 12 0.0000E+00 -6.5373E-01

11 6 7 1

12 6 8 1 REACTIONS

13 7 8 1 ---------

14 7 9 1 JOINT FORCE IN X FORCE IN Y

15 8 9 1 NUMBER DIRECTION DIRECTION

16 8 10 1 ------ ---------- ----------

17 9 10 1 1 0.000E+00 5.544E+01

18 9 11 1 11 1.251E+02 0.000E+00

19 10 11 1 12 -1.251E+02 0.000E+00

20 10 12 1

21 11 12 1

MEMBER LOADS

MEMBER PROPERTY GROUP DEFINITIONS ------------

--------------------------------- MEMBER LOAD JOINT AXIAL

PROPERTY YOUNG"S AREA OF NUMBER CASE NUMBER LOAD

GROUP # MODULUS X-SECTION ------ ---- ------ ---------

-------- --------- ---------

1 3.0000E+04 1.0000E+00 1 1 1 0.000E+00

3 0.000E+00

2 1 0.000E+00

NUMBER OF NUMBER OF 3 0.000E+00

APPLIED COMBINED 3 1 0.000E+00

LOAD CASES LOAD CASES 3 0.000E+00

---------- ---------- 4 1 0.000E+00

3 3 3 0.000E+00

5 1 0.000E+00

3 0.000E+00

6 1 0.000E+00

load case: 1 3 0.000E+00

Roof Load

2 1 1 8.520E+00

APPLIED JOINT LOADS 2 -8.520E+00

------------------- 2 1 2.434E+01

JOINT FORCE IN X FORCE IN Y 2 -2.434E+01

NUMBER DIRECTION DIRECTION 3 1 1.255E+01

------ ---------- ---------- 2 -1.255E+01

2 0.000E+00 -9.000E-01 4 1 1.193E+01

4 0.000E+00 -1.740E+00 2 -1.193E+01

6 0.000E+00 -1.680E+00 5 1 4.247E+01

8 0.000E+00 -1.680E+00 2 -4.247E+01

10 0.000E+00 -1.680E+00 6 1 5.544E+01

12 0.000E+00 -8.400E-01 2 -5.544E+01

3 1 2 -1.185E+01

3 1.185E+01

load case: 2 2 2 -3.386E+01

Snow Load 3 3.386E+01

3 2 -1.952E+01

APPLIED JOINT LOADS 3 1.952E+01

------------------- 4 2 -1.659E+01

JOINT FORCE IN X FORCE IN Y 3 1.659E+01

NUMBER DIRECTION DIRECTION 5 2 -6.239E+01

------ ---------- ---------- 3 6.239E+01

2 0.000E+00 -2.570E+00 6 2 -7.817E+01

4 0.000E+00 -4.970E+00 3 7.817E+01

6 0.000E+00 -4.800E+00

8 0.000E+00 -4.800E+00 4 1 2 9.741E+00

10 0.000E+00 -4.800E+00 4 -9.741E+00

12 0.000E+00 -2.400E+00 2 2 2.783E+01

4 -2.783E+01

3 2 1.604E+01

4 -1.604E+01

load case: 3 4 2 1.364E+01

Crane load 4 -1.364E+01

5 2 5.127E+01

APPLIED JOINT LOADS 4 -5.127E+01

------------------- 6 2 6.424E+01


NUMBER DIRECTION DIRECTION

------ ---------- ---------- 5 1 3 6.804E+00

5 0.000E+00 -6.400E+00 4 -6.804E+00

11 0.000E+00 -6.150E+00 2 3 1.944E+01

4 -1.944E+01

3 3 1.121E+01

R E S U L T OF M E S H O P T I M I Z A T I O N 4 -1.121E+01

4 3 9.525E+00

INITIAL MESH OPTIMIZED MESH 4 -9.525E+00

BANDWIDTH 2 2 5 3 3.581E+01

PROFILE 21 21 4 -3.581E+01

6 3 4.487E+01

4 -4.487E+01

OLD NODES : 1 2 3 4 5 6 7 8 9 10 11 12 6 1 3 -9.706E+00

NEW NODES : 1 2 3 4 5 6 7 8 9 10 11 12 5 9.706E+00

2 3 -2.773E+01

5 2.773E+01

SKYLINE SOLVER STATISTICS 3 3 -1.599E+01

5 1.599E+01

AVERAGE BANDWIDTH............................... 4 4 3 -1.359E+01

NUMBER OF EQUATIONS............................. 21 5 1.359E+01

NUMBER OF NON-ZERO ELEMENTS IN STIFFNESS MATRIX. 101 5 3 -5.109E+01

MAXIMUM CAPACITY OF EACH BLOCK.................. 101 5 5.109E+01


Draft9.4 †† Computer Aided Analysis 9–23

NUMBER OF BLOCK(S).............................. 1 6 3 -6.401E+01

MAXIM. CAPACITY FOR CONTROL INFORMATION......... 21 5 6.401E+01

ESTIMATE OF UNUSED STORAGE FOR EACH BLOCK....... 11

7 1 4 -7.173E+00

LENGTH OF STIFFNESS ARRAY....................... 101 5 7.173E+00

TOTAL LENGTH OF ARRAY AVAILABLE................. 49106 2 4 -2.049E+01

TOTAL NUMBER OF I/O PERFORMED................... 0 5 2.049E+01

3 4 -1.587E+01

5 1.587E+01

4 4 -1.004E+01

load case: 4 5 1.004E+01

LC 1 5 4 -4.424E+01

5 4.424E+01

6 4 -4.933E+01

NUMBER OF 5 4.933E+01

LOAD CASES

COMBINED 8 1 4 1.525E+01

---------- 6 -1.525E+01

1 2 4 4.357E+01

6 -4.357E+01

LOAD CASE COMBINATION 3 4 2.823E+01

ID NUMBER FACTOR 6 -2.823E+01

--------- ----------- 4 4 2.135E+01

1 1.400 6 -2.135E+01

5 4 8.526E+01

6 -8.526E+01

6 4 1.021E+02

load case: 5 6 -1.021E+02

LC 2

9 1 5 4.613E+00

6 -4.613E+00

NUMBER OF 2 5 1.318E+01

LOAD CASES 6 -1.318E+01

COMBINED 3 5 3.805E+00

---------- 6 -3.805E+00

3 4 5 6.459E+00

6 -6.459E+00

LOAD CASE COMBINATION 5 5 1.822E+01

ID NUMBER FACTOR 6 -1.822E+01

--------- ----------- 6 5 2.853E+01

1 1.200 6 -2.853E+01

2 0.500

3 1.600 10 1 5 -1.520E+01

7 1.520E+01

2 5 -4.342E+01

7 4.342E+01

load case: 6 3 5 -2.814E+01

LC 3 7 2.814E+01

4 5 -2.128E+01

7 2.128E+01

NUMBER OF 5 5 -8.497E+01

LOAD CASES 7 8.497E+01

COMBINED 6 5 -1.018E+02

---------- 7 1.018E+02

3

11 1 6 -3.966E+00

LOAD CASE COMBINATION 7 3.966E+00

ID NUMBER FACTOR 2 6 -1.133E+01

--------- ----------- 7 1.133E+01

1 1.200 3 6 -5.145E+00

2 1.600 7 5.145E+00

3 0.500 4 6 -5.553E+00

7 5.553E+00

5 6 -1.866E+01

7 1.866E+01

LOAD CASE: 1 6 6 -2.546E+01

7 2.546E+01

DISPLACEMENTS

------------- 12 1 6 1.818E+01

DEFLECTIONS 8 -1.818E+01

JOINT ----------------------- 2 6 5.193E+01

NUMBER DELTA-X DELTA-Y 8 -5.193E+01

------ ----------- ----------- 3 6 3.203E+01

1 -1.2412E-02 0.0000E+00 8 -3.203E+01

2 8.3395E-03 -1.2780E-03 4 6 2.545E+01

3 -1.2412E-02 -3.6280E-02 8 -2.545E+01

4 9.2804E-03 -3.7423E-02 5 6 9.902E+01

5 -1.0470E-02 -6.3848E-02 8 -9.902E+01

6 8.4821E-03 -6.4700E-02 6 6 1.209E+02

7 -7.4305E-03 -8.3526E-02 8 -1.209E+02

8 6.4357E-03 -8.4067E-02

9 -3.8080E-03 -9.4769E-02 13 1 7 2.691E+00

10 3.4968E-03 -9.4973E-02 8 -2.691E+00

11 0.0000E+00 -9.7803E-02 2 7 7.688E+00

12 0.0000E+00 -9.7640E-02 8 -7.688E+00

3 7 3.490E+00

REACTIONS 8 -3.490E+00

--------- 4 7 3.767E+00


NUMBER DIRECTION DIRECTION 5 7 1.266E+01

------ ---------- ---------- 8 -1.266E+01

1 0.000E+00 8.520E+00 6 7 1.727E+01

11 1.840E+01 0.000E+00 8 -1.727E+01

12 -1.840E+01 0.000E+00

14 1 7 -1.811E+01

9 1.811E+01

2 7 -5.175E+01

LOAD CASE: 2 9 5.175E+01

3 7 -3.192E+01

DISPLACEMENTS 9 3.192E+01

------------- 4 7 -2.536E+01

DEFLECTIONS 9 2.536E+01

JOINT ----------------------- 5 7 -9.867E+01

NUMBER DELTA-X DELTA-Y 9 9.867E+01

------ ----------- ----------- 6 7 -1.205E+02

1 -3.5460E-02 0.0000E+00 9 1.205E+02

2 2.3826E-02 -3.6510E-03

3 -3.5460E-02 -1.0365E-01 15 1 8 -1.316E+00

4 2.6515E-02 -1.0692E-01 9 1.316E+00

5 -2.9914E-02 -1.8242E-01 2 8 -3.759E+00

6 2.4234E-02 -1.8485E-01 9 3.759E+00

7 -2.1229E-02 -2.3864E-01 3 8 -4.544E+00

8 1.8387E-02 -2.4018E-01 9 4.544E+00



9 -1.0880E-02 -2.7076E-01 4 8 -1.842E+00

10 9.9907E-03 -2.7134E-01 9 1.842E+00

11 0.0000E+00 -2.7943E-01 5 8 -1.073E+01

12 0.0000E+00 -2.7896E-01 9 1.073E+01

6 8 -9.865E+00

REACTIONS 9 9.865E+00

---------

JOINT FORCE IN X FORCE IN Y 16 1 8 1.911E+01

NUMBER DIRECTION DIRECTION 10 -1.911E+01

------ ---------- ---------- 2 8 5.459E+01

1 0.000E+00 2.434E+01 10 -5.459E+01

11 5.258E+01 0.000E+00 3 8 3.524E+01

12 -5.258E+01 0.000E+00 10 -3.524E+01

4 8 2.675E+01

10 -2.675E+01

5 8 1.066E+02

LOAD CASE: 3 10 -1.066E+02

6 8 1.279E+02

DISPLACEMENTS 10 -1.279E+02

-------------

DEFLECTIONS 17 1 9 9.333E-01

JOINT ----------------------- 10 -9.333E-01

NUMBER DELTA-X DELTA-Y 2 9 2.667E+00

------ ----------- ----------- 10 -2.667E+00

1 -2.2231E-02 0.0000E+00 3 9 3.224E+00

2 1.5411E-02 -1.8825E-03 10 -3.224E+00

3 -2.2231E-02 -6.4475E-02 4 9 1.307E+00

4 1.7377E-02 -6.6357E-02 10 -1.307E+00

5 -1.9034E-02 -1.1615E-01 5 9 7.611E+00

6 1.5899E-02 -1.1685E-01 10 -7.611E+00

7 -1.3407E-02 -1.5065E-01 6 9 6.999E+00

8 1.2325E-02 -1.5136E-01 10 -6.999E+00

9 -7.0236E-03 -1.7239E-01

10 7.0396E-03 -1.7310E-01 18 1 9 -1.904E+01

11 0.0000E+00 -1.8119E-01 11 1.904E+01

12 0.0000E+00 -1.8045E-01 2 9 -5.440E+01

11 5.440E+01

REACTIONS 3 9 -3.512E+01

--------- 11 3.512E+01

JOINT FORCE IN X FORCE IN Y 4 9 -2.666E+01

NUMBER DIRECTION DIRECTION 11 2.666E+01

------ ---------- ---------- 5 9 -1.062E+02

1 0.000E+00 1.255E+01 11 1.062E+02

11 3.787E+01 0.000E+00 6 9 -1.274E+02

12 -3.787E+01 0.000E+00 11 1.274E+02

19 1 10 9.413E-01

11 -9.413E-01

LOAD CASE: 4 2 10 2.689E+00

11 -2.689E+00

DISPLACEMENTS 3 10 -4.064E+00

------------- 11 4.064E+00

DEFLECTIONS 4 10 1.318E+00

JOINT ----------------------- 11 -1.318E+00

NUMBER DELTA-X DELTA-Y 5 10 -4.028E+00

------ ----------- ----------- 11 4.028E+00

1 -1.7376E-02 0.0000E+00 6 10 3.400E+00

2 1.1675E-02 -1.7892E-03 11 -3.400E+00

3 -1.7376E-02 -5.0792E-02

4 1.2993E-02 -5.2392E-02 20 1 10 1.847E+01

5 -1.4658E-02 -8.9387E-02 12 -1.847E+01

6 1.1875E-02 -9.0580E-02 2 10 5.276E+01

7 -1.0403E-02 -1.1694E-01 12 -5.276E+01

8 9.0100E-03 -1.1769E-01 3 10 3.800E+01

9 -5.3311E-03 -1.3268E-01 12 -3.800E+01

10 4.8956E-03 -1.3296E-01 4 10 2.585E+01

11 0.0000E+00 -1.3692E-01 12 -2.585E+01

12 0.0000E+00 -1.3670E-01 5 10 1.093E+02

12 -1.093E+02

REACTIONS 6 10 1.256E+02

--------- 12 -1.256E+02

JOINT FORCE IN X FORCE IN Y

NUMBER DIRECTION DIRECTION 21 1 11 -6.936E-01

------ ---------- ---------- 12 6.936E-01

1 0.000E+00 1.193E+01 2 11 -1.982E+00

11 2.576E+01 0.000E+00 12 1.982E+00

12 -2.576E+01 0.000E+00


Draft

Chapter 10

INTERNAL FORCES INSTRUCTURES

1 This chapter will start as a review of shear and moment diagrams which you have studied in bothStatics and Strength of Materials, and will proceed with the analysis of statically determinate frames,arches and grids.

2 By the end of this lecture, you should be able to draw the shear, moment and torsion (when applicable)diagrams for each member of a structure.

3 Those diagrams will subsequently be used for member design. For instance, for flexural design, we willconsider the section subjected to the highest moment, and make sure that the internal moment is equaland opposite to the external one. For the ASD method, the basic beam equation (derived in Strength ofMaterials) σ = MC

I , (where M would be the design moment obtained from the moment diagram) wouldhave to be satisfied.

4 Some of the examples first analyzed in chapter 4 (Reactions), will be revisited here. Later on, we willdetermine the deflections of those same problems.

10.1 Design Sign Conventions

5 Before we (re)derive the Shear-Moment relations, let us arbitrarily define a sign convention.

6 The sign convention adopted here, is the one commonly used for design purposes1.

7 With reference to Fig. 10.1

2D:

Load Positive along the beam’s local y axis (assuming a right hand side convention), that ispositive upward.

Axial: tension positive.

Flexure A positive moment is one which causes tension in the lower fibers, and compression inthe upper ones. Alternatively, moments are drawn on the compression side (useful to keep inmind for frames).

Shear A positive shear force is one which is “up” on a negative face, or “down” on a positive one.Alternatively, a pair of positive shear forces will cause clockwise rotation.

Torsion Counterclockwise positive

3D: Use double arrow vectors (and NOT curved arrows). Forces and moments (including torsions) aredefined with respect to a right hand side coordinate system, Fig. ??.

1Later on, in more advanced analysis courses we will use a different one.

Draft10–2 INTERNAL FORCES IN STRUCTURES

+-

+ve Shear

+ve Load + Axial Force

+ve Moment

Figure 10.1: Shear and Moment Sign Conventions for Design

Z

Y

X

T

M

M

MT

M

x

z

y

yx

z

Figure 10.2: Sign Conventions for 3D Frame Elements


Draft10.2 Load, Shear, Moment Relations 10–3

10.2 Load, Shear, Moment Relations

8 Let us (re)derive the basic relations between load, shear and moment. Considering an infinitesimallength dx of a beam subjected to a positive load2 w(x), Fig. 10.3. The infinitesimal section must also

Figure 10.3: Free Body Diagram of an Infinitesimal Beam Segment

be in equilibrium.

9 There are no axial forces, thus we only have two equations of equilibrium to satisfy ΣFy = 0 andΣMz = 0.

10 Since dx is infinitesimally small, the small variation in load along it can be neglected, therefore weassume w(x) to be constant along dx.

11 To denote that a small change in shear and moment occurs over the length dx of the element, we addthe differential quantities dVx and dMx to Vx and Mx on the right face.

12 Next considering the first equation of equilibrium

(+ ✻)ΣFy = 0⇒ Vx + wxdx− (Vx + dVx) = 0

or

dV

dx= w(x) (10.1)

The slope of the shear curve at any point along the axis of a member is given by the loadcurve at that point.

13 Similarly

(+ �✁✛)ΣMO = 0⇒Mx + Vxdx− wxdxdx2 − (Mx + dMx) = 0

Neglecting the dx2 term, this simplifies to

dM

dx= V (x) (10.2)

The slope of the moment curve at any point along the axis of a member is given by the shearat that point.

14 Alternative forms of the preceding equations

V =∫w(x)dx (10.3)

∆V21 = Vx2 − Vx1 =∫ x2

x1

w(x)dx (10.4)

2In this derivation, as in all other ones we should assume all quantities to be positive.



The change in shear between 1 and 2, ∆V1−2, is equal to the area under the load betweenx1 and x2.

and

M =∫V (x)dx (10.5)

∆M21 = M2 −M1 =∫ x2

x1

V (x)dx (10.6)

The change in moment between 1 and 2, ∆M21, is equal to the area under the shear curvebetween x1 and x2.

15 Note that we still need to have V1 and M1 in order to obtain V2 and M2.

16 Similar relations will be determined later (Chapter 12) between curvature MEI and rotation θ (Eq.

12.30).

17 Fig. 10.4 and 10.5 further illustrates the variation in internal shear and moment under uniform andconcentrated forces/moment.

Figure 10.4: Shear and Moment Forces at Different Sections of a Loaded Beam

10.3 Moment Envelope

18 For design, we often must consider different load combinations.

19 For each load combination, we should draw the shear, moment diagrams. and then we should use theMoment envelope for design purposes.

10.4 Examples

10.4.1 Beams

Example 10-1: Simple Shear and Moment Diagram



Shear

Moment

Load

Shear

Positive Constant Negative Constant

Positive Constant Negative Constant Negative Increasing Negative Decreasing

Negative Increasing Negative Decreasing

Positive Increasing Positive Decreasing

Positive Increasing Positive Decreasing

Figure 10.5: Slope Relations Between Load Intensity and Shear, or Between Shear and Moment

Draw the shear and moment diagram for the beam shown below

Solution:The free body diagram is drawn below



Reactions are determined from the equilibrium equations

(+ ✛)ΣFx = 0; ⇒ −RAx + 6 = 0⇒ RAx = 6 k

(+ �✁✛)ΣMA = 0; ⇒ (11)(4) + (8)(10) + (4)(2)(14 + 2)−RFy(18) = 0⇒ RFy = 14 k

(+ ✻)ΣFy = 0; ⇒ RAy − 11− 8− (4)(2) + 14 = 0⇒ RAy = 13 k

Shear are determined next.

1. At A the shear is equal to the reaction and is positive.

2. At B the shear drops (negative load) by 11 k to 2 k.

3. At C it drops again by 8 k to −6 k.

4. It stays constant up to D and then it decreases (constant negative slope since the load isuniform and negative) by 2 k per linear foot up to −14 k.

5. As a check, −14 k is also the reaction previously determined at F .

Moment is determined last:

1. The moment at A is zero (hinge support).

2. The change in moment between A and B is equal to the area under the corresponding sheardiagram, or ∆MB−A = (13)(4) = 52.

3. etc...

Example 10-2: Sketches of Shear and Moment Diagrams



For each of the following examples, sketch the shear and moment diagrams.

10.4.2 Frames

20 Inclined loads on inclined members are often mishandled. With reference to Fig. 10.6 we would havethe following relations



θ

w1

w

w

2

4

L

L Y l

l l

l

l

y

x y

W

w

l

l

L X

w3

x

y

x

θ

θ

l

l

Figure 10.6: Inclined Loads on Inclined Members

w = w1lyl

+ w2lx

l+ w3

lyl

+ w4lxl

(10.7)

W = w1lylL+ w2

lx

lL+ w3

lylLY + w4

lxlLX (10.8)

lx

l= cos θ (10.9)

lyl

= sin θ (10.10)

Example 10-3: Frame Shear and Moment Diagram

Draw the shear and moment diagram of the following frame

Solution:



Reactions are determined first

(+ ✛) ΣFx = 0; ⇒ RAx −45(3)︸︷︷︸

load

(15) = 0

⇒ RAx = 36 k

(+ �✁✛)ΣMA = 0; ⇒ (3)(30)(302 ) +

35(3)(15)

(30 +

92

)︸︷︷︸

CDY

− 45(3)(15)

122︸︷︷︸

CDX

−39RDy = 0

⇒ RDy = 52.96 k

(+ ✻)ΣFy = 0; ⇒ RAy − (3)(30)− 35 (3)(15) + 52.96 = 0

⇒ RAy = 64.06 k



Shear:

1. For A−B, the shear is constant, equal to the horizontal reaction at A and negative accordingto our previously defined sign convention, VA = −36 k

2. For member B−C at B, the shear must be equal to the vertical force which was transmittedalong A−B, and which is equal to the vertical reaction at A, VB = 64.06.

3. Since B −C is subjected to a uniform negative load, the shear along B −C will have a slopeequal to −3 and in terms of x (measured from B to C) is equal to

VB−C(x) = 64.06− 3x

4. The shear along C −D is obtained by decomposing the vertical reaction at D into axial andshear components. Thus at D the shear is equal to 3

552.96 = 31.78 k and is negative. Basedon our sign convention for the load, the slope of the shear must be equal to −3 along C −D.Thus the shear at point C is such that Vc − 5

39(3) = −31.78 or Vc = 13.22. The equation forthe shear is given by (for x going from C to D)

V = 13.22− 3x

5. We check our calculations by verifying equilibrium of node C

(+ ✛)ΣFx = 0 ⇒ 35 (42.37) + 4

5 (13.22) = 25.42 + 10.58 = 36√

(+ ✻)ΣFy = 0 ⇒ 45 (42.37)− 3

5 (13.22) = 33.90− 7.93 = 25.97√

Moment:

1. Along A −B, the moment is zero at A (since we have a hinge), and its slope is equal to theshear, thus at B the moment is equal to (−36)(12) = −432 k.ft

2. Along B − C, the moment is equal to

MB−C = MB +∫ x

0

VB−C(x)dx = −432 +∫ x

0

(64.06− 3x)dx

= −432 + 64.06x− 3x2

2

which is a parabola. Substituting for x = 30, we obtain at node C: MC = −432+64.06(30)−3 30

2

2 = 139.8 k.ft

3. If we need to determine the maximum moment along B−C, we know that dMB−Cdx = 0 at the

point where VB−C = 0, that is VB−C(x) = 64.06 − 3x = 0 ⇒ x = 64.063 = 25.0 ft. In other

words, maximum moment occurs where the shear is zero.

Thus MmaxB−C = −432 + 64.06(25.0)− 3 (25.0)

2

2 = −432 + 1, 601.5− 937.5 = 232 k.ft

4. Finally along C − D, the moment varies quadratically (since we had a linear shear), themoment first increases (positive shear), and then decreases (negative shear). The momentalong C −D is given by

MC−D = MC +∫ x0 VC−D(x)dx = 139.8 +

∫ x0 (13.22− 3x)dx

= 139.8 + 13.22x− 3x2

2

which is a parabola.Substituting for x = 15, we obtain at node C MC = 139.8 + 13.22(15) − 3 15

2

2 = 139.8 +198.3− 337.5 = 0

√

Example 10-4: Frame Shear and Moment Diagram; Hydrostatic Load



The frame shown below is the structural support of a flume. Assuming that the frames are spaced 2ft apart along the length of the flume,

1. Determine all internal member end actions

2. Draw the shear and moment diagrams

3. Locate and compute maximum internal bending moments

4. If this is a reinforced concrete frame, show the location of the reinforcement.

Solution:

The hydrostatic pressure causes lateral forces on the vertical members which can be treated ascantilevers fixed at the lower end.

The pressure is linear and is given by p = γh. Since each frame supports a 2 ft wide slice of theflume, the equation for w (pounds/foot) is

w = (2)(62.4)(h)= 124.8h lbs/ft

At the base w = (124.8)(6) = 749 lbs/ft = .749 k/ft Note that this is both the lateral pressure on theend walls as well as the uniform load on the horizontal members.



End Actions

1. Base force at B is FBx = (.749)62 = 2.246 k

2. Base moment at B is MB = (2.246)63 = 4.493 k.ft

3. End force at B for member B − E are equal and opposite.

4. Reaction at C is RCy = (.749)162 = 5.99 k

Shear forces

1. Base at B the shear force was determined earlier and was equal to 2.246 k. Based on theorientation of the x− y axis, this is a negative shear.

2. The vertical shear at B is zero (neglecting the weight of A−B)

3. The shear to the left of C is V = 0 + (−.749)(3) = −2.246 k.

4. The shear to the right of C is V = −2.246 + 5.99 = 3.744 k

Moment diagrams

1. At the base: B M = 4.493 k.ft as determined above.

2. At the support C, Mc = −4.493 + (−.749)(3)(32 ) = −7.864 k.ft

3. The maximum moment is equal to Mmax = −7.864 + (.749)(5)(52 ) = 1.50 k.ft



Design: Reinforcement should be placed along the fibers which are under tension, that is on the sideof the negative moment3. The figure below schematically illustrates the location of the flexural4

reinforcement.

Example 10-5: Shear Moment Diagrams for Frame

3That is why in most European countries, the sign convention for design moments is the opposite of the one commonlyused in the U.S.A.; Reinforcement should be placed where the moment is “postive”.

4Shear reinforcement is made of a series of vertical stirrups.



Vba

Hbc

M bc Vbc

Vba

Vbc

30k 10k(10)+(2)(10)

M bc

-200’k(10)(10)+(2)(10)(10)/2

-650’k(-52.5)(12)+(-20)

450’k

(50)

(15)

-[(4

)(5)

/2][

(2)(

15)/

3)]

450’k

Hbd

Vbd

M bd

82.5k

20k

4k/ft 50k

50k

20k

(50)

-(4)

(15)

/2

M bc

12’

A

30k5k/ft

B H

M ba

8’ 10’

D

G

H

V

4k/ft

15’

D

D

AE

V

5k/ft

5’

30k

A

B C

10k

ba

20k

3.5’-22.5k

-52.5k-22.5+(-30)

17.5k

B C

17.5k

17.5-(5)(8)

-20’k

30.6’k

M ba

(17.5)(3.5)/2+(-22.5)(8-3.5)/2

(17.5)(3.5)/2

10k2k/ft

2k/ft

200’k

82.5k

450’k

52.5k 30k

00

650’k

Hbd

VbdM bd

M ba

Vba Vbc

CHECK

0

17.5-5x=0

Example 10-6: Shear Moment Diagrams for Inclined Frame



x

y

F/Fy=z/xF/Fx=z/yFx/Fy=y/x

Fy

Fx

zF

A

B

C

D

E

20k

20’36’

20’

15’

10’10’13’

13’13’

26k26k

2k/ft

125

133

5

4

H

V

a

aVe

800k

’

2k/ft60k

20k

60k

19.2k

800’k

(20)

(20)

+(6

0-20

)(20

)/2

19.2k 48.8k

48.8k

0k’20k

1130’k 488’k

+60

k+2

0k

+25.4k

-26.6k-23.1k

-39.1k800’k

60-(

2)(2

0)

(60)

(20)

-(2)

(20)

(20)

/2

26k26k

17.72k

26.6k

778k’

19.2k

0k

800k’

20k 28.8k

11.1k

-0.6-26

800’k1130’k

800+(25.4)(13)

39.1k

28.8k

778k’

0k0k’

488’k(39.1)(12.5)

-23.1-16

24k 24

k10k10k

18.46k7.38k

20-1

0-10

+25.4k

-26.6k

-23.1k-39.1k

488+

(23.

1)(1

2.5)

777k

’

1122k’

-0.58k

1,130-(.58)(13)

25.42-26

777k’

-0.58k

1122k’ 777k’

20k

16k

12k

17.28k

23.1

k

48.9k29.3k

1 2 3 4 5 6

8 B-C 11 C-D

13

1,12

2-(2

6.6)

(13)

9 B-C 12 C-D

107

AB ED

BC CD

(20)(12)/(13)=18.46(19.2)(5)/(13)=7.38(19.2)(12)/(13)=17.72

(26.6)(13)/(12)=28.8(26.6)(5)/(12)=11.1

(28.8)(3)/(5)=17.28(20)(4)/(5)=16(20)(3)/(5)=12(39.1)(5)/(4)=48.9(39.1)(3)/(4)=29.3

(28.8)(4)/(5)=23.1

(26)(12)/(13)=24

(20)(15)/13=7.7

14

17.7+7.7

7.69k

10.4.3 3D Frame

Example 10-7: 3D Frame



1. The frame has a total of 6 reactions (3 forces and 3 moments) at the support, and we have a totalof 6 equations of equilibrium, thus it is statically determinate.

2. Each member has the following internal forces (defined in terms of the local coordinate system ofeach member x′ − y′ − z′)

Member Internal ForcesMember Axial Shear Moment Torsion

Nx′ Vy′ Vz′ My′ Mz′ Tx′

C −D √ √ √ √B − C √ √ √ √ √A−B √ √ √ √ √

3. The numerical calculations for the analysis of this three dimensional frame are quite simple, howeverthe main complexity stems from the difficulty in visualizing the inter-relationships between internalforces of adjacent members.

4. In this particular problem, rather than starting by determining the reactions, it is easier to de-termine the internal forces at the end of each member starting with member C − D. Note thattemporarily we adopt a sign convention which is compatible with the local coordinate systems.

C-DΣFy′ = 0 ⇒ V Cy′ = (20)(2) = +40kNΣFz′ = 0 ⇒ V Cz′ = +60kNΣMy′ = 0 ⇒ MC

y′ = −(60)(2) = −120kN.mΣMz′ = 0 ⇒ MC

z′ = (20)(2)22 = +40kN.m

B-CΣFx′ = 0 ⇒ NB

x′ = V Cz′ = −60kNΣFy′ = 0 ⇒ V By′ = V Cy′ = +40kNΣMy′ = 0 ⇒ MB

y′ =MCy′ = −120kN.m

ΣMz′ = 0 ⇒ MBz′ = V ′

yC(4) = (40)(4) = +160kN.m

ΣTx′ = 0 ⇒ TBx′ = −MCz′ = −40kN.m



A-BΣFx′ = 0 ⇒ NA

x′ = V By′ = +40kNΣFy′ = 0 ⇒ V Ay′ = NB

x′ = +60kNΣMy′ = 0 ⇒ MA

y′ = TBx′ = +40kN.mΣMz′ = 0 ⇒ MA

z′ =MBz′ +NB

x′ (4) = 160 + (60)(4) = +400kN.mΣTx′ = 0 ⇒ TAx′ =MB

y′ = −120kN.m

The interaction between axial forces N and shear V as well as between moments M and torsionT is clearly highlighted by this example.

120 kN-m

20 kN/m

120 kN-m

40 kN

40 kN

120 kN-m40 kN

40 kN120 kN-m

40 kN40 kN

C

B

120 kN-m

120 kN-m 40 kN

120 kN-m40 kN

x’

y’

z’ x’

y’120 kN-m

120 kN-m

x’y’

60 kN40 kN-m

40 kN-m60 kN

160 kN-m40 kN-m60 kN

60 kN60 kN

40 kN-m60 kN

40 kN-m

40 kN-m

60 kN

40 kN-m60 kN

160 kN-m

160 kN-m

40 kN-m

60 kN

60 kN

160 kN-m

400 kN-m

40 kN-m

z’y’



x’

D

C

z’40

60B

Cy’ x’

z’40

B

Cy’ x’

z’160120

40

y’

x’

D

C

z’40

120

y’

A

B

x’

z’y’

60

x’

A

B

x’

z’y’

16040

400

120

A

B

V M T

M

V

T

M

V

10.5 Arches

21 See section ??.


Draft

Chapter 11

ARCHES and CURVEDSTRUCTURES

1 This chapter will concentrate on the analysis of arches.

2 The concepts used are identical to the ones previously seen, however the major (and only) differenceis that equations will be written in polar coordinates.

3 Like cables, arches can be used to reduce the bending moment in long span structures. Essentially,an arch can be considered as an inverted cable, and is transmits the load primarily through axialcompression, but can also resist flexure through its flexural rigidity.

4 A parabolic arch uniformly loaded will be loaded in compression only.

5 A semi-circular arch unifirmly loaded will have some flexural stresses in addition to the compressiveones.

11.1 Arches

6 In order to optimize dead-load efficiency, long span structures should have their shapes approximate thecoresponding moment diagram, hence an arch, suspended cable, or tendon configuration in a prestressedconcrete beam all are nearly parabolic, Fig. 20.1.

7 Long span structures can be built using flat construction such as girders or trusses. However, for spansin excess of 100 ft, it is often more economical to build a curved structure such as an arch, suspendedcable or thin shells.

8 Since the dawn of history, mankind has tried to span distances using arch construction. Essentiallythis was because an arch required materials to resist compression only (such as stone, masonary, bricks),and labour was not an issue.

9 The basic issues of static in arch design are illustrated in Fig. 20.2 where the vertical load is per unithorizontal projection (such as an external load but not a self-weight). Due to symmetry, the verticalreaction is simply V = wL

2 , and there is no shear across the midspan of the arch (nor a moment). Takingmoment about the crown,

M = Hh− wL2

(L

2− L

4

)= 0 (11.1)

Solving for H

H =wL2

8h(11.2)

We recall that a similar equation was derived for arches., and H is analogous to the C − T forces in abeam, and h is the overall height of the arch, Since h is much larger than d, H will be much smaller

Draft11–2 ARCHES and CURVED STRUCTURES

M = w L /82

L

w=W/L

BEAMBEAM

W/2 W/2

- C

+ T

- C

+ T

RISE = h

SAG = h

IDEALISTIC ARCHSHAPE GIVEN BYMOMENT DIAGRAM

IDEALISTIC SUSPENSIONSHAPE GIVEN BYMOMENT DIAGRAM

T

C

T

C

NOTE THAT THE "IDEAL" SHAPE FOR AN ARCH OR SUSPENSION

SYSTEM IS EQUIVILENT TO THE DESIGN LOAD MOMENT DIAGRAM

M-ARM smallC-T large

Figure 11.1: Moment Resisting Forces in an Arch or Suspension System as Compared to a Beam, (Linand Stotesbury 1981)

CROWN2

M = VL/2 - wL /8 - H h = 0

BASE2M = wL /8 - H h = 0

w

V = wL/2

H

w

h

L/2

h

V = wL/2

H H

VR R

L

22R = V + H

H = wL /8h2

wL/2

Figure 11.2: Statics of a Three-Hinged Arch, (Lin and Stotesbury 1981)



than C − T in a beam.

10 Since equilibrium requiresH to remain constant across thee arch, a parabolic curve would theoreticallyresult in no moment on the arch section.

11 Three-hinged arches are statically determinate structures which shape can acomodate support settle-ments and thermal expansion without secondary internal stresses. They are also easy to analyse throughstatics.

12 An arch carries the vertical load across the span through a combination of axial forces and flexuralones. A well dimensioned arch will have a small to negligible moment, and relatively high normalcompressive stresses.

13 An arch is far more efficient than a beam, and possibly more economical and aesthetic than a trussin carrying loads over long spans.

14 If the arch has only two hinges, Fig. 20.3, or if it has no hinges, then bending moments may existeither at the crown or at the supports or at both places.

APPARENT LINEOF PRESSURE WITHARCH BENDING EXCEPT AT THE BASE

h

h’

V V

M

w

h

H’=wl /8h’<2

wl /8h2

H’H’<H

APPARENT LINE OFPRESSURE WITHARCH BENDING INCLUDING BASE

V V

H

M

w

h

L

crownM

M

base

baseh’

H’<H

Figure 11.3: Two Hinged Arch, (Lin and Stotesbury 1981)

15 Since H varies inversely to the rise h, it is obvious that one should use as high a rise as possible. Fora combination of aesthetic and practical considerations, a span/rise ratio ranging from 5 to 8 or perhapsas much as 12, is frequently used. However, as the ratio goes higher, we may have buckling problems,and the section would then have a higher section depth, and the arch advantage diminishes.

16 In a parabolic arch subjected to a uniform horizontal load there is no moment. However, in practicean arch is not subjected to uniform horizontal load. First, the depth (and thus the weight) of an archis not usually constant, then due to the inclination of the arch the actual self weight is not constant.Finally, live loads may act on portion of the arch, thus the line of action will not necessarily follow thearch centroid. This last effect can be neglected if the live load is small in comparison with the dead load.

17 Since the greatest total force in the arch is at the support, (R =√V 2 +H2), whereas at the crown

we simply have H , the crown will require a smaller section than the support.


h

h’

V V

M

w

h

H’=wl /8h’<2

wl /8h2

H’H’<H


V V

H

M

w

h

L

crownM

M

base

baseh’

H’<H

Figure 11.4: Arch Rib Stiffened with Girder or Truss, (Lin and Stotesbury 1981)



11.1.1 Statically Determinate

Example 11-1: Three Hinged Arch, Point Loads. (Gerstle 1974)

Determine the reactions of the three-hinged arch shown in Fig. 20.5

Figure 11.5:

Solution:

Four unknowns, three equations of equilibrium, one equation of condition ⇒ statically determinate.

(+ �✁✛) ΣMC

z = 0; (RAy)(140) + (80)(3.75)− (30)(80)− (20)(40) +RAx(26.25) = 0⇒ 140RAy + 26.25RAx = 2.900

(+✲ )ΣFx = 0; 80−RAx −RCx = 0(+ ✻)ΣFy = 0; RAy +RCy − 30− 20 = 0

(+ �✁✛)ΣMB

z = 0; (Rax)(60)− (80)(30)− (30)(20) + (RAy)(80) = 0⇒ 80RAy + 60RAx = 3, 000

(11.3)Solving those four equations simultaneously we have:

140 26.25 0 00 1 0 11 0 1 080 60 0 0

RAyRAxRCyRCx

=

2, 9008050

3, 000

⇒

RAyRAxRCyRCx

=

15.1 k

29.8 k

34.9 k

50.2 k

(11.4)

We can check our results by considering the summation with respect to b from the right:

(+ �✁✛)ΣMB

z = 0;−(20)(20)− (50.2)(33.75) + (34.9)(60) = 0√

(11.5)

Example 11-2: Semi-Circular Arch, (Gerstle 1974)

Determine the reactions of the three hinged statically determined semi-circular arch under its owndead weight w (per unit arc length s, where ds = rdθ). 20.6Solution:

I Reactions The reactions can be determined by integrating the load over the entire structure



R cosθ

R

A

B

C

R

A

B

θ

dP=wRdθ

θ

rθ

Figure 11.6: Semi-Circular three hinged arch

1. Vertical Reaction is determined first:

(+ �✁✛) ΣMA = 0;−(Cy)(2R) +

∫ θ=π

θ=0

wRdθ︸︷︷︸dP

R(1 + cos θ)︸︷︷︸moment arm

= 0 (11.6-a)

⇒ Cy =wR

2

∫ θ=π

θ=0

(1 + cos θ)dθ =wR

2[θ − sin θ] |θ=πθ=0

=wR

2[(π − sinπ)− (0− sin 0)]

= π2wR (11.6-b)

2. Horizontal Reactions are determined next

(+ �✁✛)ΣMB = 0;−(Cx)(R) + (Cy)(R)−

∫ θ=π2

θ=0

wRdθ︸︷︷︸dP

R cos θ︸︷︷︸moment arm

= 0 (11.7-a)

⇒ Cx =π

2wR − wR

2

∫ θ=π2

θ=0

cos θdθ =π

2wR − wR[sin θ] |θ=π

2θ=0 =

π

2wR − wR(

π

2− 0)

=(π2 − 1

)wR (11.7-b)

By symmetry the reactions at A are equal to those at C

II Internal Forces can now be determined, Fig. 20.7.

1. Shear Forces: Considering the free body diagram of the arch, and summing the forces in theradial direction (ΣFR = 0):

−(π

2− 1)wR︸︷︷︸Cx

cos θ +π

2wR︸︷︷︸Cy

sin θ −∫ θ

α=0

wRdα sin θ + V = 0 (11.8)

⇒ V = wR[(π2 − 1) cos θ + (θ − π

2 ) sin θ]

(11.9)

2. Axial Forces: Similarly, if we consider the summation of forces in the axial direction (ΣFN =0):

(π2 − 1)wR sin θ + π2wR cos θ −

∫ θ

α=0

wRdα cos θ +N = 0 (11.10)

⇒ N = wR[(θ − π

2 ) cos θ − (π2 − 1) sin θ]

(11.11)

3. Moment: Now we can consider the third equation of equilibrium (ΣMz = 0):

(+ �✁✛)ΣM (π2 − 1)wR ·R sin θ − π

2wR2(1− cos θ) +∫ θ

α=0

wRdα ·R(cosα− cos θ) +M = 0 (11.12)

⇒ M = wR2[π2 (1− sin θ) + (θ − π

2 ) cos θ]

(11.13)



R cos θ

R cos αR

cos

(

)

−αθθR(1-cos )

θR

sin

C =( /2-1)wRπx

C =

/2

wR

πy

dP=wRd

VN

M

θ α

d α

α

rθ

Figure 11.7: Semi-Circular three hinged arch; Free body diagram

III Deflection are determined last

1. The real curvature φ is obtained by dividing the moment by EI

φ =M

EI=wR2

EI

[π2(1 − sin θ) + (θ − π

2) cos θ

](11.14)

1. The virtual force δP will be a unit vertical point in the direction of the desired deflection,causing a virtual internal moment

δM =R

2[1− cos θ − sin θ] 0 ≤ θ ≤ π

2(11.15)

p2. Hence, application of the virtual work equation yields:

1︸︷︷︸δP

·∆ = 2∫ π

2

θ=0

wR2

EI

[π2(1− sin θ) + (θ − π

2) cos θ

]︸︷︷︸

MEI=φ

· R2· [1− cos θ − sin θ]︸︷︷︸

δM

Rdθ︸︷︷︸dx

=wR4

16EI[7π2 − 18π − 12

]= .0337wR

4

EI (11.16-a)

11.1.2 Statically Indeterminate

Example 11-3: Statically Indeterminate Arch, (Kinney 1957)



Determine the value of the horizontal reaction component of the indicated two-hinged solid rib arch,Fig. 20.8 as caused by a concentrated vertical load of 10 k at the center line of the span. Considershearing, axial, and flexural strains. Assume that the rib is a W24x130 with a total area of 38.21 in2,that it has a web area of 13.70 in2, a moment of inertia equal to 4,000 in4, E of 30,000 k/in2, and ashearing modulus G of 13,000 k/in2.

Figure 11.8: Statically Indeterminate Arch

Solution:

1. Consider that end C is placed on rollers, as shown in Fig. ?? A unit fictitious horizontal force isapplied at C. The axial and shearing components of this fictitious force and of the vertical reactionat C, acting on any section θ in the right half of the rib, are shown at the right end of the rib inFig. 13-7.

2. The expression for the horizontal displacement of C is

1︸︷︷︸δP

∆Ch = 2∫ B

C

δMM

EIds+ 2

∫ B

C

δVV

AwGds+ 2

∫ B

C

δNN

AEds (11.17)

3. From Fig. 20.9, for the rib from C to B,

M =P

2(100−R cos θ) (11.18-a)

δM = 1(R sin θ − 125.36) (11.18-b)

V =P

2sin θ (11.18-c)

δV = cos θ (11.18-d)

N =P

2cos θ (11.18-e)



Figure 11.9: Statically Indeterminate Arch; ‘Horizontal Reaction Removed

δN = − sin θ (11.18-f)ds = Rdθ (11.18-g)

4. If the above values are substituted in Eq. 20.17 and integrated between the limits of 0.898 andπ/2, the result will be

∆Ch = 22.55 + 0.023− 0.003 = 22.57 (11.19)

5. The load P is now assumed to be removed from the rib, and a real horizontal force of 1 k is assumedto act toward the right at C in conjunction with the fictitious horizontal force of 1 k acting to theright at the same point. The horizontal displacement of C will be given by

δChCh = 2∫ B

C

δMM

EIds+ 2

∫ B

C

δVV

AwGds+ 2

∫ B

C

δNN

AEds (11.20-a)

= 2.309 + 0.002 + 0.002 = 2.313 in (11.20-b)

6. The value of the horizontal reaction component will be

HC =∆Ch

δChCh=

22.572.313

= 9.75 k (11.21)

7. If only flexural strains are considered, the result would be

HC =22.552.309

= 9.76 k (11.22)

Comments

1. For the given rib and the single concentrated load at the center of the span it is obvious that theeffects of shearing and axial strains are insignificant and can be disregarded.

2. Erroneous conclusions as to the relative importance of shearing and axial strains in the usual solidrib may be drawn, however, from the values shown in Eq. 20.19. These indicate that the effects ofthe shearing strains are much more significant than those of the axial strains. This is actually thecase for the single concentrated load chosen for the demonstration, but only because the rib doesnot approximate the funicular polygon for the single load. As a result, the shearing componentson most sections of the rib are more important than would otherwise be the case.


Draft11.2 Curved Space Structures 11–9

3. The usual arch encountered in practice, however, is subjected to a series of loads, and the axis ofthe rib will approximate the funicular polygon for these loads. In other words, the line of pressureis nearly perpendicular to the right section at all points along the rib. Consequently, the shearingcomponents are so small that the shearing strains are insignificant and are neglected.

4. Axial strains, resulting in rib shortening, become increasingly important as the rise-to-span ratioof the arch decreases. It is advisable to determine the effects of rib shortening in the design ofarches. The usual procedure is to first design the rib by considering flexural strains only, and thento check for the effects of rib shortening.

11.2 Curved Space Structures

Example 11-4: Semi-Circular Box Girder, (Gerstle 1974)

Determine the reactions of the semi-circular cantilevered box girder shown in Fig. 20.10 subjected

αd

A

y

x

z

O

R

A

B

C

y

x

z

θ

θ

rΤ ΖΜ

V

wRdα

α

Figure 11.10: Semi-Circular Box Girder

to its own weight w.Solution:

I Reactions are again determined first From geometry we have OA = R, OB = R cos θ, AB = OA −BO = R − R cos θ, and BP = R sin θ. The moment arms for the moments with respect to the xand y axis are BP and AB respectively. Applying three equations of equilibrium we obtain

FAz −∫ θ=π

θ=0

wRdθ = 0 ⇒ FAz = wRπ (11.23-a)

MAx −

∫ θ=π

θ=0

(wRdθ)(R sin θ) = 0 ⇒ MAx = 2wR2 (11.23-b)

MAy −

∫ θ=π

θ=0

(wRdθ)R(1 − cos θ) = 0 ⇒ MAy = −wR2π (11.23-c)

II Internal Forces are determined next

1. Shear Force:

(+ ✻)ΣFz = 0⇒ V −∫ θ

0

wRdα = 0⇒ V = wrθ (11.24)



2. Bending Moment:

ΣMR = 0⇒M −∫ θ

0

(wRdα)(R sinα) = 0⇒ M = wR2(1 − cos θ) (11.25)

3. Torsion:

ΣMT = 0⇒ +∫ θ

0

(wRdα)R(1 − cosα) = 0⇒ T = −wR2(θ − sin θ) (11.26)

III Deflection are determined last we assume a rectangular cross-section of width b and height d = 2band a Poisson’s ratio ν = 0.3.

1. Noting that the member will be subjected to both flexural and torsional deformations, weseek to determine the two stiffnesses.

2. The flexural stiffness EI is given by EI = E bd3

12 = E b(2b)3

12 = 2Eb4

3 = .667Eb4.

3. The torsional stiffness of solid rectangular sections J = kb3d where b is the shorter side of thesection, d the longer, and k a factor equal to .229 for d

b = 2. Hence G = E2(1+ν) = E

2(1+.3) =.385E, and GJ = (.385E)(.229b4) = .176Eb4.

4. Considering both flexural and torsional deformations, and replacing dx by rdθ:

δP∆︸︷︷︸δW

∗

=∫ π

0

δMM

EIzRdθ︸︷︷︸

Flexure

+∫ π

0

δTT

GJRdθ︸︷︷︸

Torsion︸︷︷︸δU

∗

(11.27)

where the real moments were given above.

5. Assuming a unit virtual downward force δP = 1, we have

δM = R sin θ (11.28-a)δT = −R(1− cos θ) (11.28-b)

6. Substituting these expression into Eq. 20.27

1︸︷︷︸δP

∆ =wR2

EI

∫ π

0

(R sin θ)︸︷︷︸M

(1− cos θ)︸︷︷︸δM

Rdθ +wR2

GJ

∫ π

0

(θ − sin θ)︸︷︷︸δT

R(1− cos θ)︸︷︷︸T

Rdθ

=wR4

EI

∫ π

0

[(sin θ − sin θ cos θ) +

1.265

(θ − θ cos θ − sin θ + sin θ cos θ)]dθ

=wR4

EI( 2.︸︷︷︸Flexure

+ 18.56︸︷︷︸Torsion

)

= 20.56wR4

EI (11.29-a)

11.2.1 Theory

ßAdapted from (Gerstle 1974)

18 Because space structures may have complicated geometry, we must resort to vector analysis1 todetermine the internal forces.

19 In general we have six internal forces (forces and moments) acting at any section.1To which you have already been exposed at an early stage, yet have very seldom used it so far in mechanics!



11.2.1.1 Geometry

20 In general, the geometry of the structure is most conveniently described by a parameteric set ofequations

x = f1(θ); y = f2(θ); z = f3(θ) (11.30)

as shown in Fig. 20.11. the global coordinate system is denoted by X − Y −Z, and its unit vectors are

Figure 11.11: Geometry of Curved Structure in Space

denoted2 i, j,k.

21 The section on which the internal forces are required is cut and the principal axes are identified asN − S −W which correspond to the normal force, and bending axes with respect to the Strong andWeak axes. The corresponding unit vectors are n, s,w.

22 The unit normal vector at any section is given by

n =dxi + dyj+ dzk

ds=

dxi + dyj+ dzk(dx2 + dy2 + dz2)1/2

(11.31)

23 The principal bending axes must be defined, that is if the strong bending axis is parallel to the XYplane, or horizontal (as is generally the case for gravity load), then this axis is normal to both the Nand Z axes, and its unit vector is

s =n×k|n×k| (11.32)

24 The weak bending axis is normal to both N and S, and thus its unit vector is determined from

w = n×s (11.33)

11.2.1.2 Equilibrium

25 For the equilibrium equations, we consider the free body diagram of Fig. 20.12 an applied load P2All vectorial quantities are denoted by a bold faced character.



Figure 11.12: Free Body Diagram of a Curved Structure in Space

is acting at point A. The resultant force vector F and resultant moment vector M acting on the cutsection B are determined from equilibrium

ΣF = 0; P+ F = 0; F = −P (11.34-a)ΣMB = 0; L×P+M = 0; M = −L×P (11.34-b)

where L is the lever arm vector from B to A.

26 The axial and shear forces N,Vs and Vw are all three components of the force vector F along theN,S, and W axes and can be found by dot product with the appropriate unit vectors:

N = F·n (11.35-a)Vs = F·s (11.35-b)Vw = F·w (11.35-c)

27 Similarly the torsional and bending moments T,Ms and Mw are also components of the momentvector M and are determined from

T = M·n (11.36-a)Ms = M·s (11.36-b)Mw = M·w (11.36-c)

28 Hence, we do have a mean to determine the internal forces. In case of applied loads we summ, andfor distributed load we integrate.

Example 11-5: Internal Forces in an Helicoidal Cantilevered Girder, (Gerstle 1974)

Determine the internal forces N,Vs, and Vw and the internal moments T,Ms and Mw along thehelicoidal cantilevered girder shown in FIg. 20.13 due to a vertical load P at its free end.Solution:

1. We first determine the geometry in terms of the angle θ

x = R cos θ; y = R sin θ; z =H

πθ (11.37)



Figure 11.13: Helicoidal Cantilevered Girder



2. To determine the unit vector n at any point we need the derivatives:

dx = −R sin θdθ; dy = R cos θdθ; dz =H

πdθ (11.38)

and then insert into Eq. 20.31

n =−R sin θi+R cos θj+H/πk[

R2 sin2 θ +R2 cos2 θ + (H/π)2]1/2 (11.39-a)

=1[

1 + (H/πR)2]1/2︸︷︷︸

K

[sin θi+ cos θj+ (H/πR)k] (11.39-b)

Since the denominator depends only on the geometry, it will be designated by K.

3. The strong bending axis lies in a horizontal plane, and its unit vector can thus be determined fromEq. 20.32:

n×k =1K

∣∣∣∣∣∣i j k

− sin θ cos θ HπR

0 0 1

∣∣∣∣∣∣ (11.40-a)

=1K

(cos θi+ sin θj) (11.40-b)

and the absolute magnitude of this vector |k×n| = 1K , and thus

s = cos θi+ sin θj (11.41)

4. The unit vector along the weak axis is determined from Eq. 20.33

w = s×n =1K

∣∣∣∣∣∣i j k

cos θ sin θ 0− sin θ cos θ H

πR

∣∣∣∣∣∣ (11.42-a)

=1K

(H

πRsin θi− H

πRcos θj+ k

)(11.42-b)

5. With the geometry definition completed, we now examine the equilibrium equations. Eq. 20.34-aand 20.34-b.

ΣF = 0; F = −P (11.43-a)ΣMb = 0; M = −L×P (11.43-b)

where

L = (R−R cos θ)i+ (0 −R sin θ)j+(

0− θπH

)k (11.44)

and

L×P = R

∣∣∣∣∣∣i j k

(1− cos θ) − sin θ − θπHR

0 0 P

∣∣∣∣∣∣ (11.45-a)

= PR[− sin θi− (1− cos θ)j] (11.45-b)

andM = PR[sin θi + (1− cos θ)j] (11.46)



6. Finally, the components of the force F = −Pk and the moment M are obtained by appropriatedot products with the unit vectors

N = F·n = − 1KP

HπR (11.47-a)

Vs = F·s = 0 (11.47-b)

Vw = F·w = − 1KP (11.47-c)

T = M·n = −PRK (1− cos θ) (11.47-d)

Ms = M·s = PR sin θ (11.47-e)

Mw = M·w = PHπK (1− cos θ) (11.47-f)


Draft

Chapter 12

DEFLECTION of STRUCTRES;Geometric Methods

1 Deflections of structures must be determined in order to satisfy serviceability requirements i.e. limitdeflections under service loads to acceptable values (such as ∆

L ≤ 360).

2 Later on, we will see that deflection calculations play an important role in the analysis of staticallyindeterminate structures.

3 We shall focus on flexural deformation, however the end of this chapter will review axial and torsionaldeformations as well.

4 Most of this chapter will be a review of subjects covered in Strength of Materials.

5 This chapter will examine deflections of structures based on geometric considerations. Later on, wewill present a more pwerful method based on energy considerations.

12.1 Flexural Deformation

12.1.1 Curvature Equation

6 Let us consider a segment (between point 1 and point 2), Fig. 12.1 of a beam subjected to flexuralloading.

7 The slope is denoted by θ, the change in slope per unit length is the curvature κ, the radius ofcurvature is ρ.

8 From Strength of Materials we have the following relations

ds = ρdθ ⇒ dθ

ds=

1ρ

(12.1)

9 We also note by extension that ∆s = ρ∆θ

10 As a first order approximation, and with ds ≈ dx and dydx = θ Eq. 12.1 becomes

κ =1ρ

=dθ

dx=d2y

dx2(12.2)

11 † Next, we shall (re)derive the exact expression for the curvature. From Fig. 12.1, we have

tan θ =dy

dx(12.3)

Draft12–2 DEFLECTION of STRUCTRES; Geometric Methods

Figure 12.1: Curvature of a flexural element

Defining t as

t =dy

dx(12.4)

and combining with Eq. 12.3 we obtainθ = tan−1 t (12.5)

12 Applying the chain rule to κ = dθds we have

κ =dθ

dt

dt

ds(12.6)

ds can be rewritten asds =

√dx2 + dy2

=

√1 +

(dydx

)2dx

t = dydx

ds =

√1 + t2dx (12.7)

13 Next combining Eq. 12.6 and 12.7 we obtain

κ = dθdt

dt√1+t2dx

θ = tan−1 tdθdt = 1

1+t2

κ = 1

1+t21√1+t2

dtdx

dtdx = d2y

dx2

κ =

d2ydx2[

1 +(dydx

)2] 32 (12.8)

14 Thus the slope θ, curvature κ, radius of curvature ρ are related to the y displacement at a point xalong a flexural member by

κ =d2ydx2[

1 +(dydx

)2] 32 (12.9)


Draft12.1 Flexural Deformation 12–3

15 If the displacements are very small, we will have dydx << 1, thus Eq. 12.9 reduces to

κ =d2y

dx2=

1ρ

(12.10)

12.1.2 Differential Equation of the Elastic Curve

16 Again with reference to Figure 12.1 a positive dθ at a positive y (upper fibers) will cause a shorteningof the upper fibers

∆u = −y∆θ (12.11)

17 This equation can be rewritten as

lim∆s→0

∆u∆s

= −y lim∆s→0

∆θ∆s

(12.12)

and since ∆s ≈ ∆xdu

dx︸︷︷︸ε

= −y dθdx

(12.13)

Combining this with Eq. 12.10

1ρ

= κ = − εy

(12.14)

This is the fundamental relationship between curvature (κ), elastic curve (y), and linear strain (ε).

18 Note that so far we made no assumptions about material properties, i.e. it can be elastic or inelastic.

19 For the elastic case:εx = σ

E

σ = −MyI

}ε = −My

EI(12.15)

Combining this last equation with Eq. 12.14 yields

1ρ

=dθ

dx=d2y

dx2=M

EI(12.16)

This fundamental equation relates moment to curvature.

20 Combining this equation with the moment-shear-force relations determined in the previous chapter

dVdx = w(x)dMdx = V (x)

}d2M

dx2(12.17-a)

we obtain

w(x)EI

=d4y

dx4(12.18)

12.1.3 Moment Temperature Curvature Relation

21 Assuming linear variation in temperature

∆T = α∆TTdx Top (12.19-a)∆B = α∆TBdx Bottom (12.19-b)



22 Next, consideringd2y

dx2=M

EI= − ε

y(12.20)

In this case, we can take ε = α∆TT at y = h2 or ε = α(TT − TB) at y = h thus

dθ

dx=d2y

dx2=M

EI= −α(TT − TB)

h(12.21)

12.2 Flexural Deformations

12.2.1 Direct Integration Method

23 Equation 12.18 lends itself naturally to the method of double integration which was presented inStrength of Materials

Example 12-1: Double Integration

Determine the deflection at B for the following cantilevered beam

Solution:

At: 0 ≤ x ≤ 2L3

1. Moment Equation

EId2y

dx2=Mx =

wL

3x− 5

18wL2 (12.22)

2. Integrate once

EIdy

dx=wL

6x2 − 5

18wL2x+ C1 (12.23)

However we have at x = 0, dydx = 0, ⇒ C1 = 0

3. Integrate twice

EIy =wL

18x3 − 5wL2

36x2 + C2 (12.24)

Again we have at x = 0, y = 0, ⇒ C2 = 0


Draft12.2 Flexural Deformations 12–5

At: 2L3 ≤ x ≤ L

1. Moment equation

EId2y

dx2=Mx =

wL

3x− 5

18wL2 − w(x − 2L

3)(x − 2L

3

2) (12.25)

2. Integrate once

EIdy

dx=wL

6x2 − 5

18wL2x− w

6(x − 2L

3)3 + C3 (12.26)

Applying the boundary condition at x = 2L3 , we must have dy

dx equal to the value coming fromthe left, ⇒ C3 = 0

3. Integrating twice

EIy =wL

18x3 − 5

36wL2x2 − w

24(x − 2L

3)4 + C4 (12.27)

Again following the same argument as above, C4 = 0

Substituting for x = L we obtain

y =1631944

wL4

EI(12.28)

12.2.2 Curvature Area Method (Moment Area)

12.2.2.1 First Moment Area Theorem

24 From equation 12.16 we havedθ

dx=M

EI(12.29)

this can be rewritten as (note similarity with dVdx = w(x)).

θ21 = θ2 − θ1 =∫ x2

x1

dθ =∫ x2

x1

M

EIdx (12.30)

or with reference to Figure 12.2

First Area Moment Theorem: The change in slope from point 1 to point 2 on a beam isequal to the area under the M/EI diagram between those two points.

12.2.2.2 Second Moment Area Theorem

25 Similarly, with reference to Fig. 12.2, we define by t21 the distance between point 2 and the tangentat point 1. For an infinitesimal distance ds = ρdθ and for small displacements

dt = dθ(x2 − x)dθdx = M

EI

}dt =

M

EI(x2 − x)dx (12.31)

To evaluate t21

t21 =∫ x2

x1

dt =∫ x2

x1

M

EI(x2 − x)dx (12.32)

or

Second Moment Area Theorem: The tangent distance t21 between a point, 2, on thebeam and the tangent of another point, 1, is equal to the moment of the M/EI diagrambetween points 1 and 2, with respect to point 2.



Figure 12.2: Moment Area Theorems



Figure 12.3: Sign Convention for the Moment Area Method

Figure 12.4: Areas and Centroid of Polynomial Curves



26 The sign convention is as shown in Fig. 12.3

27 Fig. 12.4 is a helpful tool to determine centroid and areas.

Example 12-2: Moment Area, Cantilevered Beam

Determine the deflection of point A

Solution:

EItA/C =12

(−2wL2

5

)(5L2

)︸︷︷︸

area

(23

5L2

)

︸︷︷︸Moment wrt A

+13

(−wL22

)(L)︸︷︷︸

area

(9L4

)

︸︷︷︸Moment wrt A

= −29wL4

24(12.33)

Thus, ∆A = 29wL4

24EI

Example 12-3: Moment Area, Simply Supported Beam

Determine ∆C and θC for the following example



Solution:

Deflection ∆C is determined from ∆C = c′c = c′c”− c”c, c”c = tC/B, and c′c” = tA/B2 from geometry

tA/B =1EI

(

3Pa4

)(a2

)︸︷︷︸

A1

(2a3

)

︸︷︷︸Moment

+(

3Pa4

)(3a2

)︸︷︷︸

A2+A3

(a+

3a3

)

︸︷︷︸Moment

=5Pa3

2EI(12.34)



Figure 12.5: Maximum Deflection Using the Moment Area Method

This is positive, thus above tangent from B

tC/B =1EI

(Pa

2

)(2a2

)(2a3

)=Pa3

3EI(12.35)

Positive, thus above the tangent from B Finally,

∆C =5Pa3

4EI− Pa

3

3EI=

1112Pa3

EI(12.36)

Rotation θC is

θBC = θB − θC ⇒ θC = θB − θBCθBC = A3

θB = tA/BL

⇒ θC =

5Pa3

2EI(4a)−(Pa

2

)(2a2

)=Pa2

8EI(12.37)

12.2.2.3 Maximum Deflection

28 A joint along the beam will have the maximum (or minimum) relative deflection if dydx = 0 or θ = 0

So we can determine ∆max if x is known, Fig. 12.5.To determine x:

1. Compute tC/A

2. θA = tC/AL

3. θB = 0 and θAB = θA − θB, thus θAB = θA. Hence, compute θAB in terms of x using FirstTheorem.

4. Equate items 2 and 3, then solve for x.

Example 12-4: Maximum Deflection



Determine the deflection at D, and the maximum deflection at B

Solution:

Deflection at D:

∆D = tD/A − 95tC/A (12.38-a)

EItC/A =12

(−4PL5

)(L)

(L

3

)(12.38-b)

tC/A = −2PL3

15EI(12.38-c)

EItD/A =12

(−4PL5

)(L)

(17L15

)+

12

(−4PL5

)(4L5

)(8L15

)(12.38-d)

tD/A = −234375

PL3

EI(12.38-e)

Substituting we obtain

∆D = − 48125

PL3

EI(12.39)

Maximum deflection at B:

tC/A = −2PL3

15EI(12.40-a)

θA =tC/A

L= −2PL2

15EI(12.40-b)

θAB =1EI

[12

(4Px5

)(x)]

= −25Px2

EI(12.40-c)

θA = θAB ⇒ − 215PL2

EI= −2

5Px2

EI⇒ x =

L√3

(12.40-d)

∆max =(

4Px5EI

)(x2

)(23x

)at x =

L√3

(12.40-e)

⇒ ∆max =4PL3

45√

3EI(12.40-f)

Example 12-5: Frame Deflection



Complete the following example problem

Solution:xxxx

Example 12-6: Frame Subjected to Temperature Loading

Neglecting axial deformation, compute displacement at A for the following frame

Solution:

1. First let us sketch the deformed shape



2. BC flexes ⇒ θB = θC �= 0

3. Rigid hinges at B and C with no load on AB and CD

4. Deflection at A

∆A = AA′′ = AA′ +A′A′′ (12.41-a)AA′ = ∆B = ∆C = |θC |h2 (12.41-b)A′′A′ = |θB|h1 (12.41-c)

5. We need to compute θB & θC

θB =tC/BL

(12.42-a)

θC = θCB + θB or θC =tB/CL

(12.42-b)

6. In order to apply the curvature are theorem, we need a curvature (or moment diagram).

1ρ

= α(TB − TT

h) =

M

EI(12.43)

7.

tC/B = A(L

2

)⇒ |θB| = A

(L

2

)(1L

)=A

2or θB = −A

2(-ve) (12.44)

8.θCB = AθC = θCB + θB

}θC = A− A

2=A

2(+ve) (12.45)

9. From above,

∆A = |θC |h2 + |θB|h1 = A2 h2 + A

2 h1= A

2 (h2 + h1)A = α(TB−TT )

h L

∆A = α(TB − TT )

L

h

(12

)(h2 + h1) (12.46)

10. Substitute

∆A = (6.5× 10−6)(200− 60)(20)(12)

(16)(12)(10 + 25)(12) (12.47-a)

= 2.87 in (12.47-b)

11. Other numerical values:

θB = θC =A

2=

12α(TB − TT )

hL (12.48-a)

=12(6.5× 10−6)

(200− 60)(16)

(20)(12) (12.48-b)

= .00683rad. (.39 degrees) =dy

dx<< 1 (12.48-c)

Note: sin(.00683) = .006829947 and tan(.00683) = .006830106

M

EI=

1ρ

= α(TB − TT )

h(12.49-a)

= (6.5× 10−6)(200− 60)

16= 5.6875× 10−5 (12.49-b)

⇒ ρ =1

5.6875× 10−5= 1.758× 104 in = 1, 465 ft (12.49-c)



12. In order to getM , we need E & I. Note the difference with other statically determinate structures;the stiffer the beam, the higher the moment; the higher the moment, the higher the stress? NO!!

13. σ = MyI = EI

ρyI = Ey

ρ

14. ρ is constant ⇒ BC is on arc of circle M is constant & MEI = d2y

dx2 ⇒ d2ydx2 = M

EI = C ⇒ y =C x2

2 + dx+ e

15. The slope is a parabola, (Why?)

1ρ

=M

EI=

d2ydx2

(1 + (

dy

dx)2)

32 ≈ d

2y

dx2(12.50)

29 Let us get curvature from the parabola slope and compare it with ρ

y =cx2

2+ dx+ e (12.51-a)

dy

dx= cx+ d (12.51-b)

d2y

dx2= c (12.51-c)

at x = 0, y = 0, thus e = 0at x = 0, dyx = θB = −.00683 thus d = −.00683at x = 20 ft, dy

dx = θC = .00683 thus c(20)− .00683 = .00683 thus c = 6.83× 10−4 thus

y = 6.83× 10−4(x2

2− 10x

)(12.52-a)

dy

dx= 6.83× 10−4(x− 10) (12.52-b)

d2y

dx2= 6.83× 10−4 (12.52-c)

κ = 6.83× 10−4 (12.52-d)

ρ =1

6.83× 10−4= 1, 464ft as expected! (12.52-e)

If we were to use the exact curvature formula

d2ydx2

1 + ( dydx)2)32 =

6.83× 10−4

[1 + (.00683)2]32

(12.53-a)

= 6.829522× 10−4 (12.53-b)⇒ ρ = 1464.23ft (compared with 1465ft) (12.53-c)

12.2.3 Elastic Weight/Conjugate Beams

V and M θ and y

V12 =

∫ x2

x1

wdx V =

∫wdx + C1 θ12 =

∫ x2

x1

1

ρdx θ =

∫1

ρdx + C1

M12 =

∫ x2

x1

V dx M =

∫V dx + C2 t21 =

∫ x2

x1

θdx y =

∫θdx + C2



Figure 12.6: Conjugate Beams

29 This leads to the following

Load q ≡ curvature1ρ

= κ =M

EI(12.54)

Shear V ≡ slope θ (12.55)Moment M ≡ deflection y (12.56)

30 Since V & M can be conjugated from statics, by analogy θ & y can be thought of as the V & M ofa fictitious beam (or conjugate beam) loaded by M

EI elastic weight.

31 What about Boundary Conditions? Table 12.1, and Fig. 12.6.

Actual Beam Conjugate BeamHinge θ �= 0 y = 0 V �= 0 M = 0 “Hinge”Fixed End θ = 0 y = 0 V = 0 M = 0 Free endFree End θ �= 0 y �= 0 V �= 0 M �= 0 Fixed endInterior Hinge θ �= 0 y �= 0 V �= 0 M �= 0 Interior supportInterior Support θ �= 0 y = 0 V �= 0 M = 0 Interior hinge

Table 12.1: Conjugate Beam Boundary Conditions

32 Whereas the Moment area method has a well defined basis, its direct application can be sometimesconfusing.

33 Alternatively, the moment area method was derived from the moment area method, and is a farsimpler method to remember and use in practice when simple “back of the envelope” calculations arerequired.

34 Note that we can only have distributed load, and that the load the load is positive for a positivemoment, and negative for a negative moment. “Shear” and “Moment” diagrams should be drawnaccordingly.

35 Units of the “distributed load” w∗ are FLEI (force time length divided by EI). Thus the “Shear” would

have units of w∗ × L or FL2

EI and the “moment” would have units of (w∗ × L) × L or FL3

EI . Recallingthat EI has units of FL−2L4 = FL2, we observe that indeed the “shear” corresponds to a rotation inradians and the “moment” to a displacement.



Example 12-7: Conjugate Beam

Analyze the following beam.

Solution:

3 equations of equilibrium and 1 equation of condition = 4 = number of reactions.Deflection at D = Shear at D of the corresponding conjugate beam (Reaction at D)Take AC and ΣM with respect to C

RA(L)−(

4PL5EI

)(L

2

)(L

3

)= 0 (12.57-a)

⇒ RA =2PL2

15EI(12.57-b)

(Slope in real beam at A) As computed before!Let us draw the Moment Diagram for the conjugate beam

M =P

EI

[215L2x−

(45x

)(x2

)(x3

)](12.58-a)

=P

EI

(215L2x− 2

15x3)

(12.58-b)

=2P

15EI(L2x− x3) (12.58-c)

Point of Maximum Moment (∆max) occurs when dMdx = 0

dM

dx=

2P15EI

(L2 − 3x2) = 0⇒ 3x2 = L2 ⇒ x =L√3

(12.59)


Draft12.3 Axial Deformations 12–17

Figure 12.7: Torsion Rotation Relations

as previously determined

x =L√3

(12.60-a)

⇒M =2P

15EI

(L2L√

3− L3

3√

3

)(12.60-b)

=4PL3

45√

3EI(12.60-c)

as before.

12.3 Axial Deformations

Statics: σ = PA

Material: σ = Eε

Kinematics: Σ = ∆L

∆ =PL

AE(12.61)

12.4 Torsional Deformations

36 Since torsional effects are seldom covered in basic structural analysis, and students may have forgottenthe derivation of the basic equations from the Strength of Material course, we shall briefly review thebasic equations.

37 Assuming a linear elastic material, and a linear strain (and thus stress) distribution along the radiusof a circular cross section subjected to torsional load, Fig. 12.7 we have:

T =∫A

ρ

cτmax︸︷︷︸stress

dA︸︷︷︸area︸︷︷︸

Force

ρ︸︷︷︸arm

︸︷︷︸torque

(12.62)

=τmaxc

∫A

ρ2dA︸︷︷︸J

(12.63)

τmax =Tc

J(12.64)



Note the analogy of this last equation with σ = McIy

38

∫A

ρ2dA is the polar moment of inertia J for circular cross sections and is equal to:

J =∫A

ρ2dA =∫ c

0

ρ2 (2πρdρ)

=πc4

2=πd4

32(12.65)

39 Having developed a relation between torsion and shear stress, we now seek a relation between torsionand torsional rotation. Considering Fig. 12.7-b, we look at the arc length BD

γmaxdx = dΦc⇒ dΦdx = γmax

cγmax = τmax

G

}dΦdx = τmax

Gc

τmax = TCJ

dΦdx =

T

GJ(12.66)

40 Finally, we can rewrite this last equation as∫Tdx =

∫GjdΦ and obtain:

T = GJL Φ (12.67)


Draft

Chapter 13

ENERGY METHODS; Part I

13.1 Introduction

1 Energy methods are powerful techniques for both formulation (of the stiffness matrix of an element1)and for the analysis (i.e. deflection) of structural problems.

2 We shall explore two techniques:

1. Real Work

2. Virtual Work (Virtual force)

13.2 Real Work

3 We start by revisiting the first law of thermodynamics:

The time-rate of change of the total energy (i.e., sum of the kinetic energy and the internalenergy) is equal to the sum of the rate of work done by the external forces and the changeof heat content per unit time.

ddt (K + U) =We +H (13.1)

where K is the kinetic energy, U the internal strain energy, We the external work, and H the heat inputto the system.

4 For an adiabatic system (no heat exchange) and if loads are applied in a quasi static manner (nokinetic energy), the above relation simplifies to:

We = U (13.2)

5 Simply stated, the first law stipulates that the external work must be equal to the internal strainenergy due to the external load.

13.2.1 External Work

6 The external work is given by, Fig. 13.1

We =∫ ∆f

0

Pd∆

=∫ θf

0

Mdθ

(13.3)1More about this in Matrix Structural Analysis.

Draft13–2 ENERGY METHODS; Part I

��

��

��

��

K

dW

K1

M

θ

P

∆dW

K1

PPM M

∆ θ

Figure 13.1: Load Deflection Curves

for point loads and concentrated moments respectively.

7 For linear elastic systems, we have for point loads

P = K∆

We =∫ ∆f

0

Pd∆

We = K

∫ ∆f

0

∆d∆ =12K∆2

f (13.4)

When this last equation is combined with Pf = K∆f we obtain

We =12Pf∆f (13.5)

where K is the stiffness of the structure.

8 Similarly for an applied moment we have

We =12Mfθf (13.6)

13.2.2 Internal Work

9 Considering an infinitesimal element from an arbitrary structure subjected to uniaxial state of stress,the strain energy can be determined with reference to Fig. 13.2. The net force acting on the elementwhile deformation is taking place is P = σxdydz. The element will undergo a displacement u = εxdx.Thus, for a linear elastic system, the strain energy density is dU = 1

2σε. And the total strain energywill thus be

U =12

∫Volε Eε︸︷︷︸

σ

dVol (13.7)

10 When this relation is applied to various structural members it would yield:


Draft13.2 Real Work 13–3

Figure 13.2: Strain Energy Definition

Axial Members:U =

∫Vol

εσ

2dVol

σ = PA

ε = PAE

dV = Adx

U =

∫ L

0

P 2

2AEdx (13.8)

Torsional Members:

U = 12


σ

dVol

U = 12

∫Volγxy Gγxy︸︷︷︸

τxy

dVol

τxy = TrJ

γxy = τxyG

dVol = rdθdrdx

J =∫ r

o

∫ 2π

0

r2dθ dr

U =∫ L

0

T 2

2GJdx (13.9)

Flexural Members:U = 1

2


σ

σx = MzyIz

ε = MzyEIz

dVol = dAdx

Iz =∫A

y2dA

U =

∫ L

0

M2

2EIzdx (13.10)

Example 13-1: Deflection of a Cantilever Beam, (Chajes 1983)

Determine the deflection of the cantilever beam, Fig. 13.3 with span L under a point load P appliedat its free end. Assume constant EI.Solution:

We =12P∆f (13.11-a)



��

��

��

P

M=PL

x

Figure 13.3: Deflection of Cantilever Beam

U =∫ L

0

M2

2EIdx (13.11-b)

M = −Px (13.11-c)

U =P 2

2EI

∫ L

0

x2dx =P 2L3

6EI(13.11-d)

12P∆f =

P 2L3

6EI(13.11-e)

∆f =PL3

3EI(13.11-f)

13.3 Virtual Work

11 A severe limitation of the method of real work is that only deflection along the externally appliedload can be determined.

12 A more powerful method is the virtual work method.

13 The principle of Virtual Force (VF) relates force systems which satisfy the requirements of equilibrium,and deformation systems which satisfy the requirement of compatibility

14 In any application the force system could either be the actual set of external loads dp or somevirtual force system which happens to satisfy the condition of equilibrium δp. This set of externalforces will induce internal actual forces dσ or internal virtual forces δσ compatible with the externallyapplied load.

15 Similarly the deformation could consist of either the actual joint deflections du and compatible internaldeformations dε of the structure, or some virtual external and internal deformation δu and δε whichsatisfy the conditions of compatibility.

16 It is often simplest to assume that the virtual load is a unit load.

17 Thus we may have 4 possible combinations, Table 13.1: where: d corresponds to the actual, and δ(with an overbar) to the hypothetical values.

This table calls for the following observations

1. The second approach is the same one on which the method of virtual or unit load is based. It issimpler to use than the third as a internal force distribution compatible with the assumed virtual


Draft13.3 Virtual Work 13–5

Force Deformation IVW FormulationExternal Internal External Internal

1 dp dσ du dε

2 δp δσ du dε δU∗

Flexibility3 dp dσ δu δε δU Stiffness4 δp δσ δu δε

Table 13.1: Possible Combinations of Real and Hypothetical Formulations

force can be easily obtained for statically determinate structures. This approach will yield exactsolutions for statically determinate structures.

2. The third approach is favored for statically indeterminate problems or in conjunction with approx-imate solution. It requires a proper “guess” of a displacement shape and is the basis of the stiffnessmethod.

18 Let us consider an arbitrary structure and load it with both real and virtual loads in the followingsequence, Fig. 13.4. For the sake of simplicity, let us assume (or consider) that this structure developsonly axial stresses.

P 1

∆1

P 1

∆1

��

��

��

��

P

ε

σ

∆1

��

��

��

��

δ ∆

δε

σδ

δP

σδ δPε = ∆

Real

Virtual

Inter. Ext.

��

��

��

��

δP

σδ

ε

∆

εδ

δ ∆

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

Virtual System

ε

∆

σ

δP

ε

σδ

δ∆+∆δP

δ ∆

δε

σδ

Virtual + Real LoadsReal Load

-

σδ

δP P

ε

σ

∆1

Figure 13.4: Real and Virtual Forces

1. If we apply the virtual load, then

12δPδ∆ =

12

∫dVol

δσδεdVol (13.12)



2. Load with the real (applied) load, since the external work must be equal to the internal strainenergy over the entire volume, then:

12P1∆1 =

12

∫dVol

σεdVol (13.13)

3. We now immagine that the virtual load was first applied, and we then apply the real (actual) loadon top of it, then the total work done is

12δPδ∆ +

12P1∆1︸︷︷︸+ δP∆︸︷︷︸ =

12

∫VolδσδεdVol +

12

∫VolσεdVol︸︷︷︸+

∫VolδσεdVol︸︷︷︸ (13.14)

4. Since the strain energy and work done must be the same whether the loads are applied togetheror separately, we obtain, from substracting the sum of Eqs. 13.13 and 13.12 from 13.14 andgeneralizing, we obtain ∫

(δσε+ δτγ) dVol︸︷︷︸δU

∗

= δP∆︸︷︷︸δW

∗(13.15)

19 This last equation is the key to the method of virtual forces. The left hand side is the internal virtualstrain energy δU

∗2. Similarly the right hand side is the external virtual work.

13.3.1 External Virtual Work δW∗

20 The general expression for the external virtual work δW∗

is

δW∗

=∑n

i=1(∆i)δP i +

∑n

i=1(θi)δM i (13.16)

for distributed load, point load, and concentrated moment respectively.

21 Recall that all overbar quantities are virtual and the other ones are the real.

13.3.2 Internal Virtual Work δU∗

22 The general expression for the internal virtual work is

∫(δσε+ δτγ) dVol︸︷︷︸

δU∗

= δP∆︸︷︷︸δW

∗(13.17)

We will generalize it to different types of structural members.

23 We will first write the equations independently of the material stress strain relation, and then we willrewrite those same equations for a linear elastic system.

General :

Axial Members:

δU∗

=∫ L

0

δσεdVol

dVol = Adx

δU

∗= A

∫ L

0

δσεdx (13.18)



Figure 13.5: Torsion Rotation Relations

Torsional Members: With reference to Fig. 13.5 Note that in torsion the “strain” is γ the rateof change of rotation of the cross section about the longitudinal axis x.

δU∗

=∫

Volδτxyγxy dVol

δT =∫A

δτxyrdA

γxy = r dθdxdVol = dAdx

δU

∗=∫ L

0

(∫A

δτxyrdA)︸︷︷︸δT

θdx

⇒ δU∗

=∫ L

0

δTdθ

dxdx

(13.19)

Shear Members:

δU∗

=∫

Volδτxyγxy dVol

δV =∫A

δτxydA

dVol = dAdx

δU

∗=∫ L

0

(∫A

δτxydA)︸︷︷︸δV

γxydx

⇒ δU∗

=∫ L

0

δV γxydx

(13.20)

Flexural Members:

δU∗

=∫δσxεxdVol

δM =∫A

δσxydA⇒ δM

y=∫A

δσxdA

φ = εy

φy = ε

dVol =∫ L

0

∫A

dAdx

δU

∗=∫ L

0

δMφdx (13.21)

Linear Elastic Systems Should we have a linear elastic material

σ = Eε (13.22)

then the previous equations can be rewritten as:2We use the * to distinguish it from the internal virtual strain energy obtained from the virtual displacement method

δU .



Axial Members:δU

∗=

∫VolεδσdVol

δσ = δPA

ε = PAE

dV = Adx

δU

∗=∫ L

0

δPP

AE︸︷︷︸∆

dx (13.23)

Note that for a truss where we have n members, the above expression becomes

δU∗

= Σn1 δP iPiLiAiEi

(13.24)

Shear Members:

δU∗

=∫

VolδτxyγxydVol

τxy = k VAγxy = τxy

GdVol = Adx

λdef=

∫A

k2dydz

δU

∗= λ

∫ L

0

δVV

GA︸︷︷︸γxy

dx (13.25)

24 Note that the exact expression for the shear stress is

τ =V Q

Ib(13.26)

where Q is the moment of the area from the external fibers to y with respect to the neutralaxis; For a rectangular section, this yields

τ =V Q

Ib(13.27-a)

=V

Ib

∫ h/2

y

by′dy′ =V

2I

(h2

4− y2

)(13.27-b)

=6Vbh3

(h2

4− 4)

(13.27-c)

and we observe that the shear stress is zero for y = h/2 and maximum at the neutral axiswhere it is equal to 1.5 Vbh .

25 To determine the form factor λ of a rectangular section

τ = V QIb

= k VA

Q =∫ h/2

y

by′dy′ =b

2

(h2

4− y2

)

k = A2I

(h2

4 − y2)

λ bh︸︷︷︸A

=∫A

k2dydz

λ = 1.2 (13.28)

26 Thus, the form factor λ may be taken as 1.2 for rectangular beams of ordinary proportions.

27 For I beams, k can be also approximated by 1.2, provided A is the area of the web.

Torsional Members:

δU∗

=∫

Volδτxy

τxyG︸︷︷︸γxy

dVol

τxy = TrJ

γxy = τxyG

dVol = rdθdrdx

J =∫ r

o

∫ 2π

0

r2dθ dr

δU

∗=∫ L

0

δT︸︷︷︸δσ

T

GJ︸︷︷︸ε

dx (13.29)



28 Note the similarity with the corresponding equation for shear deformation.

29 The torsional stiffness of cylindrical sections is given by J = πd4

32 .

30 The torsional stiffness of solid rectangular sections is given by

J = kb3d (13.30)

where b is the shorter side of the section, d the longer, and k a factor given by Table 13.2.

d/b 1.0 1.5 2.0 2.5 3.0 4.0 5.0 10 ∞k 0.141 0.196 0.229 0.249 0.263 0.281 0.291 0.312 0.333

Table 13.2: k Factors for Torsion

31 Recall from strength of materials that

G =E

2(1 + ν)(13.31)

Flexural Members:

δU∗

=∫

Volε Eδε︸︷︷︸

δσ

dV ol

σx = MzyIz

ε = MzyEIz

dVol = dAdx

Iz =∫A

y2dA

δU

∗=∫ L

0

δMM

EIz︸︷︷︸Φ

dx (13.32)

13.3.3 Examples

Example 13-2: Beam Deflection (Chajes 1983)

Determine the deflection at point C in Fig. 13.6 E = 29, 000 ksi, I = 100 in4.Solution:

For the virtual force method, we need to have two expressions for the moment, one due to the realload, and the other to the (unit) virtual one, Fig. 13.7.

Element x = 0 M δMAB A 15x− x2 −0.5xBC C −x2 −x

Applying Eq. 13.32 we obtain

∆CδP︸︷︷︸δW

∗

=∫ L

0

δMM

EIzdx︸︷︷︸

δU∗

(13.33-a)

(1)∆C =∫ 20

0

(−0.5x)(15x− x2)

EIdx +

∫ 10

0

(−x)−x2

EIdx (13.33-b)



Figure 13.6:

15x -x

-x-0.5x

2

Figure 13.7:.



=2, 500EI

(13.33-c)

∆C =(2, 500) k2 − ft3(1, 728) in3/ ft3

(29, 000) ksi(100) in4(13.33-d)

= 1.49 in (13.33-e)

Example 13-3: Deflection of a Frame (Chajes 1983)

Determine both the vertical and horizontal deflection at A for the frame shown in Fig. 13.8. E =200× 106 kN/ m2, I = 200× 106 mm4.

Figure 13.8:.

Solution:

To analyse this frame we must determine analytical expressions for the moments along each memberfor the real load and the two virtual ones. One virtual load is a unit horizontal load at A, and the othera unit vertical one at A also, Fig. 13.9.

Element x = 0 M δMv δMh

AB A 0 x 0BC B 50x 2 + x 0CD C 100 4 −x

Note that moments are considered positive when they produce compression on the inside of the frame.



+100

50x +

+4

-x

x x

x

-

+

+

50 kN1 kN

1 kN

4 kN-m100 kN-m 5 kN-m

50 kN 1 kN

1 kN

Figure 13.9:

Substitution yields:

∆vδP︸︷︷︸δW

∗

=∫ L

0

δMM

EIzdx︸︷︷︸

δU∗

(13.34-a)

(1)∆v =∫ 2

0

(x)(0)EIdx+

∫ 2

0

(2 + x)50xEI

dx +∫ 5

0

(4)100EIdx (13.34-b)

=2, 333 kN m3

EI(13.34-c)

=(2, 333) kN m3(103)4 mm4/ m4

(200× 106) kN/ m2(200× 106) mm4(13.34-d)

= 0.058 m = 5.8 cm (13.34-e)

Similarly for the horizontal displacement:

∆hδP︸︷︷︸δW

∗

=∫ L

0

δMM

EIzdx︸︷︷︸

δU∗

(13.35-a)

(1)∆h =∫ 2

0

(0)(0)EIdx+

∫ 2

0

(0)50xEI

dx+∫ 5

0

(−x)100EIdx (13.35-b)

=−1, 250 kN m3

EI(13.35-c)

=(−1, 250) kN m3(103)4 mm4/ m4

(200× 106) kN/ m2(200× 106) mm4(13.35-d)

= −0.031 m = −3.1 cm (13.35-e)

Example 13-4: Rotation of a Frame (Chajes 1983)

Determine the rotation of joint C for frame shown in Fig. 13.10. E = 29, 000 ksi, I = 240 in4.Solution:

In this problem the virtual force is a unit moment applied at joint C, δMe. It will cause an internalmoment δM i



Figure 13.10:

Element x = 0 M δMAB A 0 0BC B 30x− 1.5x2 −0.05xCD D 0 0

Note that moments are considered positive when they produce compression on the outside of the frame.Substitution yields:

θCδM e︸︷︷︸δW

∗

=∫ L

0

δMM

EIzdx︸︷︷︸

δU∗

(13.36-a)

(1)θC =∫ 20

0

(−0.05x)(30x− 1.5x2)

EIdx k2 ft3 (13.36-b)

=(1, 000)(144)(29, 000)(240)

(13.36-c)

= −0.021 radians (13.36-d)

Example 13-5: Truss Deflection (Chajes 1983)

Determine the vertical deflection of joint 7 in the truss shown in Fig. 13.11. E = 30, 000 ksi.Solution:

Two analyses are required. One with the real load, and the other using a unit vertical load at joint7. Results for those analysis are summarized below. Note that advantage was taken of the symmetricload and structure.

Member A L P δP δPPLA n nδPPL

Ain2 ft k k

1 & 4 2 25 -50 -0.083 518.75 2 1,037.510 & 13 2 20 40 0.67 268.0 2 536.011 & 12 2 20 40 0.67 268.0 2 536.05 & 9 1 15 20 0 0 2 06 & 8 1 25 16.7 0.83 346.5 2 693.02& 3 2 20 -53.3 -1.33 708.9 2 1,417.8

7 1 15 0 0 0 1 0Total 4,220.3



Figure 13.11:

Thus from Eq. 13.24 we have:

∆δP︸︷︷︸δW

∗

=∫ L

0

δPP

AEdx︸︷︷︸

δU∗

(13.37-a)

= ΣδPPL

AE(13.37-b)

1∆ =(4, 220.3)(12)

30, 000(13.37-c)

= 1.69 in (13.37-d)

Example 13-6: Torsional and Flexural Deformation, (Chajes 1983)

Determine the vertical deflection at A in the structure shown in Fig. 13.12. E = 30, 000 ksi, I =144 in4, G = 12, 000 ksi, J = 288 in4

Solution:

1. In this problem we have both flexural and torsional deformation. Hence we should determine theinternal moment and torsion distribution for both the real and the unit virtual load.2. Then we will use the following relation

δW∗︷︸︸︷

δP∆A =

δU∗︷︸︸︷∫

δMM

EIdx︸︷︷︸

flexure

+∫δT

T

GJdx︸︷︷︸

Torsion

3. The moment and torsion expressions are given by



Figure 13.12:

Element x = 0 M δM T δTAB A 10x x 0 0BC B 15x x 50 5

4. Substituting,

δP∆A =∫δM

M

EIdx+

∫δT

T

GJdx

1∆A =∫ 5

0

x10xEI

dx+∫ 5

0

x15xEIdx+

∫ 5

0

(5)50GJdx

= (1,042)(1,728)(30,000)(144) + (1,250)(1,728)

(12,000)(288)

= 0.417 + 0.625= 1.04 in

Example 13-7: Flexural and Shear Deformations in a Beam (White et al. 1976)

Determine the delection of a cantilevered beam, of length L subjected to an end force P due to bothflexural and shear deformations. Assume G = 0.4E, and a square solid beam cross section.Solution:

1. The virtual work equation is

δP︸︷︷︸1

·∆ =∫ L

0

δMdφ+∫ L

0

δV γxydx (13.38-a)

=∫ L

0

δMM

EIdx+ λ

∫ L

0

δV λV

GAdx (13.38-b)

2. The first integral yields for M = Px, and δM = (1)(x)∫ L

0

δMM

EIdx =

P

EI

∫ L

0

x2dx (13.39-a)

= PL3/3EI (13.39-b)



3. The second integral represents the contribution of the shearing action to the total internal virtualwork and hence to the total displacement.4. Both the real shear V and virtual shear δV are constant along the length of the member, hence∫ L

0

δV λV

GAdx =

λ

GA

∫ L

0

1(P )dx =λPL

GA(13.40)

5. Since λ= 1.2 for a square beam; hence

1 ·∆ =PL3

3EI+

1.2PLGA

(13.41-a)

=PL

3E

[L2

I+

3.60.4A

]= PL

3E

[L2

I + 9A

](13.41-b)

6. For a square beam of dimension h

I =h4

12and A = h2 (13.42)

then

∆ =PL

3E

[12L2

h4+

9h2

]=

3PLEh2

[1.33L2

h2+ 1]

(13.43)

7. Choosing L = 20 ft and h = 1.5 ft (L/h = 13.3)

∆ =3PLEh2

[1.33(

201.5

)2+ 1] = 3PL

Eh2 (237 + 1) (13.44)

Thus the flexural deformation is 237 times the shear displacement. This comparison reveals why wenormally neglect shearing deformation in beams.

As the beam gets shorter or deeper, or as L/h decreases, the flexural deformation decreases relativeto the shear displacement. At L/h = 5, the flexural deformation has reduced to 1.33(5)2 = 33 times theshear displacement.

Example 13-8: Thermal Effects in a Beam (White et al. 1976)

The cantilever beam of example 13-7 is subjected to a thermal environment that produces a temper-ature change of 70◦C at the top surface and 230◦C at the bottom surface, Fig. 13.13.

If the beam is a steel, wide flange section, 2 m long and 200 mm deep, what is the angle of rotation,θ1, at the end of beam as caused by the temperature effect? The original uniform temperature of thebeams was 40◦C.Solution:

1. The external virtual force conforming to the desired real displacement θ1 is a moment δM = 1 at thetip of the cantilever, producing an external work term of moment times rotation. The internal virtualforce system for this cantilever beam is a uniform moment δM int = 1.2. The real internal deformation results from: (a) the average beam temperature of 150◦C, which is110◦C above that of the original temperature, and (b) the temperature gradient of 160◦C across thedepth of the beam.3. The first part of the thermal effect produces only a lengthening of the beam and does not enter intothe work equation since the virtual loading produces no axial force corresponding to an axial change inlength of the beam.4. The second effect (thermal gradient) produces rotation dφ, and an internal virtual work term of∫ L0 δMdφ.



Figure 13.13:

5. We determine the value of dφ by considering the extreme fiber thermal strain as shown above. Theangular rotation in the length dx is the extreme fiber thermal strain divided by half the beam depth, or

dφ =ε

h/2=α∆Tdxh/2

=(11.7× 10−6)(80)dx

100=

936× 10−6

100= (9.36× 10−6)dx (13.45)

6. Using the virtual work equation

1 · θ1 =∫ L

0

δMdφ (13.46-a)

=∫ 2,000

0

1(9.36× 10−6)dx (13.46-b)

= +0.01872 (13.46-c)

= 0.01872 �✁✛ radians (13.46-d)

7. This example raises the following points:

1. The value of θ1 would be the same for any shape of 200 mm deep steel beam that has its neutralaxis of bending at middepth.

2. Curvature is produced only by thermal gradient and is independent of absolute temperature values.

3. The calculation of rotations by the method of virtual forces is simple and straightforward; theapplied virtual force is a moment acting at the point where rotation is to be calculated.

4. Internal angular deformation dφ has been calculated for an effect other than load-induced stresses.The extension of the method of virtual forces to treat inelastic displacements is obvious – all weneed to know is a method for determining the inelastic internal deformations.

Example 13-9: Deflection of a Truss (White et al. 1976)



12’ 12’

12’

11

3

4

4

7

5

2 3

5

60 k 120 k

Figure 13.14:

Determine the deflection at node 2 for the truss shown in Fig. 13.14.Solution:

δP P , L, A, E, δP PLAE

Member kips kips ft in2 ksi1 +0.25 +37.5 12 5.0 10× 103 +22.5× 10−4

2 +0.25 +52.5 12 5.0 10× 103 +31.5× 10−4

3 -0.56 -83.8 13.42 5.0 10× 103 +125.9× 10−4

4 +0.56 +16.8 13.42 5.0 10× 103 +25.3× 10−4

5 +0.56 -16.8 13.42 5.0 10× 103 −25.3× 10−4

6 -0.56 -117.3 13.42 5.0 10× 103 +176.6× 10−4

7 -0.50 -45.0 12 5.0 10× 103 +54.0× 10−4

+410.5× 10−4

The deflection is thus given by

δP∆ =7∑1

δPPL

AE(13.47-a)

∆ = (410.5× 10−4)(12 in/ ft) = 0.493 in (13.47-b)

Example 13-10: Thermal Defelction of a Truss; I (White et al. 1976)



The truss in example 13-9 the preceding example is built such that the lower chords are shielded fromthe rays of the sun. On a hot summer day the lower chords are 30◦ F cooler than the rest of the trussmembers. What is the magnitude of the vertical displacement at joint 2 as a result of this temperaturedifference?Solution:

1. The virtual force system remains identical to that in the previous example because the desired dis-placement component is the same.2. The real internal displacements are made up of the shortening of those members of the truss that areshielded from the sun.3. Both bottom chord members 1 and 2 thus shorten by

∆l = α(∆T )(L) = (0.0000128) in/ in/oF (30)o(12) ft(12) in/ft = 0.0553 in (13.48)

4. Then,

1 ·∆ =∑

δP (∆L) = .25(−0.0553)+ .25(−0.0553) = −0.0276 (13.49-a)

= −0.0276 in ✻ (13.49-b)

5. The negative sign on the displacement indicates that it is in opposite sense to the assumed directionof the displacement; the assumed direction is always identical to the direction of the applied virtualforce.6. Note that the same result would be obtained if we had considered the internal displacements to bemade up of the lengthening of all truss members above the bottom chord.

Example 13-11: Thermal Deflections in a Truss; II (White et al. 1976)

A six-panel highway bridge truss, Fig. 13.15 is constructed with sidewalks outside the trusses so that

Figure 13.15:

the bottom chords are shaded. What will be the vertical deflection component of the bottom chord atthe center of the bridge when the temperature of the bottom chord is 40◦F (∆T ) below that of the topchord, endposts, and webs? (coefficient of steel thermal expansion is α=0.0000065 per degree F.)Solution:

1. The deflection is given by

∆δP︸︷︷︸δW

∗

=∫ L

0

δPP

AEdx︸︷︷︸

δU∗

= ΣδPPL

AE= ΣδP∆L = ΣδPα∆TL (13.50)



where ∆L is the temperature change in the length of each member, and δP are the member virtualinternal forces.2. Taking advantage of symmetry:

(0.0000065)(40)LMember L (ft) α∆TL δP (k) δP∆L

4 35 +0.00910 + 0.625 +0.005685 21 +0.00546 + 0.75 +0.004096 21 +0.00546 +1.13 +0.006167 0 0 0 08 35 +0.00910 -0.625 -0.005689 28 +0.00728 +0.5 +0.0036410 35 +0.00910 -0.625 -0.0056811 0 0 0 0

+0.00821

3. Hence, the total deflection is

∆ = (2)(0.00821)(12) in/ft = +0.20 in ✻ (13.51)

4. A more efficient solution would have consisted in considering members 1,2, and 3 only and apply a∆T = −40, we would obtain the same displacement.5. Note that the forces in members 1, 2, and 3 (-0.75, -0.375, and -0.375 respectively) were not includedin the table because the corresponding ∆T = 0.6. A simpler solution would have ∆T = −40 in members 1, 2, and 3 thus,

(0.0000065)(40)LMember L (ft) α∆TL δP (k) δP∆L

1 21 -0.00546 -0.75 +0.0040952 21 -0.00546 -0.375 +0.00204753 21 -0.00546 -0.375 +0.0020475

+0.00819

∆ = (2)(0.00819)(12) in/ft = +0.20 in ✻ (13.52)

Example 13-12: Truss with initial camber

It is desired to provide 3 in. of camber at the center of the truss shown below

by fabricating the endposts and top chord members additionally long. How much should the length ofeach endpost and each panel of the top chord be increased?Solution:

1. Assume that each endpost and each section of top chord is increased 0.1 in.



Member δP int ∆L δP int∆L1 +0.625 +0.1 +0.06252 +0.750 +0.1 +0.07503 +1.125 +0.1 +0.1125

+0.2500

Thus,(2)(0.250) = 0.50 in (13.53)

2. Since the structure is linear and elastic, the required increase of length for each section will be(3.00.50

)(0.1) = 0.60 in (13.54)

3. If we use the practical value of 0.625 in., the theoretical camber will be

(6.25)(0.50)0.1

= 3.125 in (13.55)

Example 13-13: Prestressed Concrete Beam with Continously Variable I (White et al. 1976)

A prestressed concrete beam is made of variable depth for proper location of the straight pretensioningtendon, Fig. 13.16. Determine the midspan displacement (point c) produced by dead weight of the girder.The concrete weights 23.6 kN/m3 and has E = 25, 000 MPa (N/mm2). The beam is 0.25 m wide.Solution:

1. We seek an expression for the real momentM , this is accomplished by first determining the reactions,and then considering the free body diagram.2. We have the intermediary resultant forces

R1 = (0.25)x(0.26)m3(23.6)kN/m3 = 3.54x (13.56-a)R2 = frac12(0.25)x(0.04x)m3(23.6)kN/m3 = 0.118x2 (13.56-b)

Hence,

M(x) = 47.2x− 3.54x(x

2

)− 0.118x2

(x3

)(13.57-a)

= 47.2x− 1.76x2 − 0.0393x3 (13.57-b)

3. The moment of inertia of the rectangular beam varies continously and is given, for the left half of thebeam, by

I(x) =112bh3 =

112

(0.25)︸︷︷︸148

(0.6 + 0.04x)3 (13.58)

4. Thus, the real angle changes produced by dead load bending are

dφ =M

EIdx =

47.2x− 1.76x2 − 0.0393x3

E(148

)(0.6 + 0.04x)3

dx (13.59)

5. The virtual force system corresponding to the desired displacement is shown above with δM = (1/2)xfor the left half of the span. Since the beam is symmetrical, the virtual work equations can be evaluatedfor only one half of the beam and the final answer is then obtained by multiplying the half-beam resultby two.



Figure 13.16: *(correct 42.7 to 47.2)



6. The direct evaluation of the integral∫δMdφ is difficult because of the complexity of the expression for

dφ. Hence we shall use a numerical procedure, replacing the∫δM(M/EI)dx with the

∑δM(M/EI)∆x,

where each quantity in the summation is evaluated at the center of the interval ∆x and held constantover the interval length. As ∆x becomes very short, the solution approaches the exact answer.7. An interval length of 1 meter, giving 10 elements in the half length of the beam, is chosen to establishan accurate result.8. The internal virtual work quantity is then∫ L/2

0

δMM

EIdx

∑ δMM

EI(∆x) (13.60-a)

=∑ δMM

E(0.2512

)h3

(∆x) (13.60-b)

48E

∑ δMM

h3(∆x) (13.60-c)

9. The summation for the 10 elements in the left half of the beam gives

Segment x h h3 M δM δMMh3

1 0.5 0.62 0.238 23.2 0.25 242 1.5 0.66 0.288 66.7 0.75 1743 2.5 0.70 0.343 106.4 1.25 3884 3.5 0.74 0.405 150 1.75 6485 4.5 0.78 0.475 173 2.25 8206 5.5 0.82 0.551 200 2.75 9987 6.5 0.86 0.636 222 3.25 1,1348 7.5 0.90 0.729 238 3.75 1,2249 8.5 0.94 0.831 250 4.25 1,279

10 9.5 0.98 0.941 256 4.75 1,2927,981

10. The SI units should be checked for consistency. Letting the virtual force carry the units of kN, thevirtual moment δM has the units of m·kN, and the units of the equation

∆c =1

1 kN

∑ δMM

EI∆x (13.61)

are1kN

( m · kN)( m · kN)( MN/ m2) m4

m =m

1, 000= mm (13.62)

11. Then ∫ L

0

δMM

EIdx 2

[48

25, 000(7, 981)(1)

]= 30.6mm (13.63)

and the deflection at midspan is∆c = 30.6 mm (13.64)

12. Acceptably accurate results may be obtained with considerably fewer elements (longer intervals ∆x).Using four elements with centers at 2, 5, 8, and 10, the

∑(δMM/h3)∆x is∑

= 3(174) + 3(820) + 3(1, 224) + 1(1, 292) = 7, 946 (13.65)

which is only 0.4% lower than the 10 element solution. If we go to two elements, 3 and 8, we obtain asummation of 5(388) + 5(1, 224) = 8, 060, which is 1% high. A one element solution, with x = 5 m andh = 0.8 m, gives a summation of 9,136, which is 14.4% high and much less accurate than the 2 elementsolution.13. Finally, it should be noted that the calculations involved in this example are essentially identical tothose necessary in the moment area method.



13.4 *Maxwell Betti’s Reciprocal Theorem

24 If we consider a beam with two points, A, and B, we seek to determine

1. The deflection at A due to a unit load at B, or fAB

2. The deflection at B due to a unit load at A, or fBA

25 Applying the theorem of vitual work

fAB =∫ L

0

δMAMB

EIdx (13.66-a)

fBA =∫ L

0

δMBMA

EIdx (13.66-b)

But since the both the real and the virtual internal moments are caused by a unit load, both momentsare numerically equal

δMA = MA (13.67-a)δMB = MB (13.67-b)

thus we conclude that

fBA = fBA (13.68)

or The displacement at a point B on a structure due to a unit load acting at point A is equal to thedisplacement of point A when the unit load is acting at point B.

26 Similarly The rotation at a point B on a structure due to a unit couple moment acting at point A isequal to the rotation at point A when a unit couple moment is acting at point B.

27 And The rotation in radians at point B on a structure due to a unit load acting at point A is equalto the displacement of point A when a unit couple moment is acting at point B.

28 These theorems will be used later on in justifying the symmetry of the stiffness matrix, and inconstruction of influence lines using the Muller-Breslau principle.

13.5 Summary of Equations


Draft13.5 Summary of Equations 13–25

U Virtual Force δU∗

General Linear

Axial∫ L

0

P 2

2AEdx

∫ L

0

δσεdx

∫ L

0

δP︸︷︷︸δσ

P

AE︸︷︷︸ε

dx

Shear ...

∫ L

0

δV γxydx ...

Flexure∫ L

0

M2

2EIzdx

∫ L

0

δMφdx

∫ L

0

δM︸︷︷︸δσ

M

EIz︸︷︷︸ε

dx

Torsion∫ L

0

T 2

2GJdx

∫ L

0

δTθdx

∫ L

0

δT︸︷︷︸δσ

T

GJ︸︷︷︸ε

dx

W Virtual Force δW∗

P Σi 12Pi∆i ΣiδP i∆i

M Σi 12Miθi ΣiδM iθi

w

∫ L

0

w(x)v(x)dx∫ L

0

δw(x)v(x)dx

Table 13.3: Summary of Expressions for the Internal Strain Energy and External Work


Draft

Chapter 14

BRACED ROLLED STEEL BEAMS

1 This chapter deals with the behavior and design of laterally supported steel beams according to theLRFD provisions.

2 If a beam is not laterally supported, we will have a failure mode governed by lateral torsional buckling.

3 By the end of this lecture you should be able to select the most efficient section (light weight withadequate strength) for a given bending moment and also be able to determine the flexural strength of agiven beam.

14.1 Review from Strength of Materials

14.1.1 Flexure

4 Fig.14.1 shows portion of an originally straight beam which has been bent to the radius ρ by endcouples M , thus the segment is subjected to pure bending. It is assumed that plane cross-sectionsnormal to the length of the unbent beam remain plane after the beam is bent. Therefore, consideringthe cross-sections AB and CD a unit distance apart, the similar sectors Oab and bcd give

ε =y

ρ(14.1)

where y is measured from the axis of rotation (neutral axis). Thus strains are proportional to thedistance from the neutral axis.

5 The corresponding variation in stress over the cross-section is given by the stress-strain diagram ofthe material rotated 90o from the conventional orientation, provided the strain axis ε is scaled with thedistance y (Fig.14.1). The bending moment M is given by

M =∫yσdA (14.2)

where dA is an differential area a distance y (Fig.14.1) from the neutral axis. Thus the moment M canbe determined if the stress-strain relation is known.

6 If stress is proportional to strain, σ = Eε, Equations 14.1 and 14.2 give

M = Eρ

∫y2dA =

EI

ρ= EIε

y = σIy

= σS

where

S =I

y(14.3)

Draft14–2 BRACED ROLLED STEEL BEAMS

D

C

a bMM

O

ρ

B

Ac d y

Figure 14.1: Bending of a Beam

is the Elastic Section Modulus. Note that for doubly symmetric sections, we have two section modulusSx and Sy.

7 The stress distribution on a typical wide-flange shape subjected to increasing bending moment is shownin Fig.14.2. In the service range (that is before we multiplied the load by the appropriate factors in theLRFD method) the section is elastic. This elastic condition prevails as long as the stress at the extremefiber has not reached the yield stress Fy . Once the strain ε reaches its yield value εy, increasing straininduces no increase in stress beyond Fy.

Figure 14.2: Stress distribution at different stages of loading

8 When the yield stress is reached at the extreme fiber, the nominal moment strength Mn, is referredto as the yield moment My and is computed as

Mn =My = SxFy (14.4)

(assuming that bending is occurring with respect to the x− x axis).

9 When across the entire section, the strain is equal or larger than the yield strain (ε ≥ εy = Fy/Es)then the section is fully plastified, and the nominal moment strength Mn is therefore referred to as the


Draft14.1 Review from Strength of Materials 14–3

Elastic Region Plastic Region

Stress

Strain ε

F

Fy

Figure 14.3: Stress-strain diagram for most structural steels

plastic moment Mp and is determined from

Mp = Fy∫A

ydA = FyZ (14.5)

Z =∫ydA (14.6)

is the Plastic Section Modulus.

10 The ratio Mp/My is a property of the cross-sectional shape and is independent of the material prop-erties, it is referred to as the Shape Factor

ξ =Mp

My=Z

S(14.7)

The shape factors of some regular cross-sectional shapes bending about their major axis are:

1. Rectangular section = 1.50

2. I section = 1.11

3. Circular section = 1.70

14.1.2 Shear

11 Examining the stress distribution in FIg. 14.4, if there is a variation of moment along the infinitesimalsegment dx (thus a shear force is present), we determine the total force acting over the shaded area onthe left of the section dx, on the right, and then take the difference which must be equal to the shearforce along the top surface of the shaded zone

σxdA =My

IdA (14.8-a)

F1 =∫A

My

IdA (14.8-b)

F2 =∫A

(M + dM)yI

dA (14.8-c)

τbdx = F2 − F1 (14.8-d)



��

��

τmax

��

��

h/2

h/2 M

y

z

M+dM

y1

dx

b

h

dA

y

τ

O

σx

Figure 14.4: Flexural and Shear Stress Distribution in a Rectangular Beam

=∫A

dMy

IdA (14.8-e)

τ =dM

dx

(1Ib

)∫A

ydA (14.8-f)

(14.8-g)

substituting V = dM/dx we now obtain

τ =V Q

Ib(14.9)

Q =∫A

ydA (14.10)

this is the shear formula, and Q is the first moment of the area.

12 For a rectangular beam, we will have a parabolic shear stress distribution.

13 For a W section, it can be easily shown that about 95% of the shear force is carried by the web, andthat the average shear stress is within 10% of the actual maximum shear stress.

14.2 Nominal Strength

The strength requirement for beams in load and resistance factor design is stated as

φbMn ≥Mu (14.11)

where:φb strength reduction factor; for flexure 0.90Mn nominal moment strengthMu factored service load moment.

14 The equations given in this chapter are valid for flexural members with the following kinds of crosssection and loading:

1. Doubly symmetric (such as W sections) and loaded in Plane of symmetry

2. Singly symmetric (channels and angles) loaded in plane of symmetry or through the shear centerparallel to the web1.

1More about shear centers in Mechanics of Materials II.


Draft14.3 Flexural Design 14–5

14.3 Flexural Design

14.3.1 Failure Modes and Classification of Steel Beams

15 The strength of flexural members is limited by:

Plastic Hinge: at a particular cross section.

local buckling: of a cross-sectional element (e.g. the web or the flange),

Lateral-Torsional buckling: of the entire member.

Fig. 14.5 illustrates some of the buckling mode possible in a rolled section.

Figure 14.5: Local (flange) Buckling; Flexural and Torsional Buckling Modes in a Rolled Section, (LuleaUniversity)

16 Accordingly, the LRFD manual classifies steel sections as

Compact sections: No local buckling can occur. Strength is based on the plastic moment.

Partially compact sections: Where local buckling may occur

Slender sections: where lateral torsional buckling may occur.

We will cover only the first two cases.

14.3.1.1 Compact Sections

17 For compact sections, the mode of failure is the formation of a plastic hinge that is the section isfully plastified. Hence we shall first examine the bending behavior of beams under limit load. Then wewill relate this plastic moment to the design of compact sections.

18 A section is compact if the following conditions are met:

1. Flanges are continuously connected to the web

2. Width to thickness ratios, known as the slenderness ratios, of the flange and the web must notexceed the limiting ratios λp defined as follows:

Flange bf2tf≤ λp λp = 65√

Fy

Web hctw≤ λp λp = 640√

Fy

(14.12)

Note that bf2tf

and hctw

are tabulated in Sect. 3.6.



19 The nominal strength Mn for laterally stable compact sections according to LRFD is

Mn =Mp (14.13)

where:Mp plastic moment strength = ZFyZ plastic section modulusFy specified minimum yield strength

20 Note that section properties, including Z values are tabulated in Section 3.6.

14.3.1.2 Partially Compact Section

21 If the width to thickness ratios of the compression elements exceed the λp values mentioned in Eq.14.12 but do not exceed the following λr, the section is partially compact and we can have local buckling.

Flange: λp <bf2tf≤ λr λp = 65√

Fyλr = 141√

Fy−FrWeb: λp <

hctw≤ λr λp = 640√

Fyλr = 970√

Fy

(14.14)

where, Fig. 14.6:Fy specified minimum yield stress in ksi

bf width of the flangetf thickness of the flangehc unsupported height of the web which is twice the distance from the neutral axis

to the inside face of the compression flange less the fillet or corner radius.tw thickness of the web.Fr residual stress = 10.0 ksi for rolled sections and 16.5 ksi for welded sections.

Figure 14.6: W Section

22 The nominal strength of partially compact sections according to LRFD is, Fig. 14.7

Mn =Mp − (Mp −Mr)(λ− λpλr − λp ) ≤Mp (14.15)

where:Mr Residual Moment equal to (Fy − Fr)Sλ bf/2tf for I-shaped member flanges and

hc/tw for beam webs.



bf

hc

Fy

Fy

Fy

Fr

M

M

M

n

p

r

λλ λp r

Flanges2t

f

Webt

w

141

640

65

-Fy

970

Compact Partially Compact Slender

Figure 14.7: Nominal Moments for Compact and Partially Compact Sections

23 All other quantities are as defined earlier. Note that we use the λ associated with the one beingviolated (or the lower of the two if both are).

14.3.1.3 Slender Section

24 If the width to thickness ratio exceeds λr values of flange and web, the element is referred to as slendercompression element. Since the slender sections involve a different treatment, it will not be dealt here.

25 A laterally stable beam is one which is braced laterally in the direction perpendicular to the planeof the web. Thus overall buckling of the compression flange as a column cannot occur prior to its fullparticipation to develop the moment strength of the section.

26 The AISC code requires that

Lb ≤ Lp (14.16)

Lp =300√Fyrmin (14.17)

where Lb is the maximum unbraced length of the compression flange, and rmin is the radius of gyrationwith respect to the weak axis. This is an emperical equation, thus units are in inches, ksi, and ft.

14.3.2 Examples

Example 14-1: Shape Factors, Rectangular Section

Determine the shape factor for a rectangular beam of width b and depth d.



Solution:From Eq. 14.7 the shape factor is given byξ = Mp

MyThe moment at first yield is

My =∫A

fydA

f = Fyyd/2 = Fy 2yd

⇒My = 2

∫ d/2

0

2Fydy2bdy = Fy

bd2

6

The plastic moment is given by:

Mp =∫A

fydA = 2∫ d/2

0

Fybydy = Fybd2

4

Thus, the shape factor is

ξ =Mp

My=Fy

bd2

4

Fybd2

6

= 1.5

Example 14-2: Shape Factors, T Section

Determine the shape factor of the inverted T-section shown in fig. below.

Solution:

1. Determine the position of the Elastic Neutral Axis:

Taking moment about the top, the depth of the neutral axis is given by

y =(10)(1)(9.5) + (9)(1)(4.5)

(10)(1) + (9)(1)= 7.13 cm



2. Determine the moment of inertia about the elastic neutral axis:

I =[(1)(9)3

12+ (9)(1)(7.13− 4.5)2

]+[(10)(1)3

12+ (10)(1) ((10− 7.13)− 0.5)2

]= 180 cm4

3. Determine My:

My = FyI

ymax= Fy

180.07.13

= 25.25Fy

4. Determine the position of Plastic Neutral Axis:

Plastic neutral axis divides the section in such a way that, the area above it is equal to area belowit. In other words it is also called as the bf Equal Area Axis

Total area = (10)(1) + (9)(1) = 19 cm2 Half area = 9.5 cm2

Thus the plastic neutral axis passes through the horizontal seat at 0.95 cm. from the bottom.

5. Determine Mp: Taking moments of the resultant forces C1 and C2 with respect to the location ofthe resultant force of the lower tensile forces we have:

Mp = [(Fy)(1)(9)][4.5 + 0.05 +0.952

] + [(Fy)(10)(0.05)][0.052

+0.952

] = 45.475Fy

6. Shape Factor

ξ =Mp

My=

45.475Fy25.25Fy

= 1.8

Example 14-3: Beam Design

Select the lightest W or M section to carry a uniformly distributed dead load of 0.2 kip/ft superim-posed (i.e., in addition to the beam weight) and 0.8 kip/ft live load. The simply supported span is 20ft. The compression flange of the beam is fully supported against lateral movement. Select the sectionsfor the following steels: A36; A572 Grade 50; and A572 Grade 65.Solution:

Case 1: A36 Steel

1. Determine the factored load.

wD = 0.2 k/ft

wL = 0.8 k/ft

wu = 1.2wD + 1.6wL= 1.2(0.2) + 1.6(0.8) = 1.52 k/ft

2. Compute the factored load moment Mu. For a simply supported beam carrying uniformlydistributed load,

Mu = wuL2/8 = (1.52)(20)2/8 = 76 k.ft

Assuming compact section, since a vast majority of rolled sections satisfy λ ≤ λp for both theflange and the web. The design strength φbMn is

φbMn = φbMp = φbZxFy

The design requirement isφbMn =Mu

or, combing those two equations we have:

φbZxFy =Mu



3. Required Zx is

Zx =Mu

φbFy=

76(12)0.90(36)

= 28.1 in3

From the notes on Structural Materials, we select a W12X22 section which has a Zx = 29.3 in3

Note that Zx is approximated by wd9 = (22)(12)

9 = 29.3.

4. Check compact section limits λp for the flanges from the table

λ = bf2tf

= 4.7λp = 65√

Fy= 65√

36= 10.8 > λ

√

and for the web:λ = hc

tw= 41.8

λp = 640√Fy

= 640√36

= 107√

5. Check the Strength by correcting the factored moment Mu to include the self weight. Selfweight of the beam W12X22 is 22 lb./ft. or 0.022 kip/ft

wD = 0.2 + 0.022 = 0.222 k/ft

wu = 1.2(0.222) + 1.6(0.8) = 1.55 k/ft

Mu = (1.55)(20)2/8 = 77.3 k.ft

Mn = Mp = ZxFy = (29.3) in3(36) ksi(12) in/ft

= 87.9 k.ft

φbMn = 0.90(87.9) = 79.1 k.ft > Mu√

Therefore use W12X22 section.

6. We finally check for the maximum distance between supports.

rmin =

√IyA

=

√5

6.5= 0.88 in (14.18-a)

Lp =300√Fyrmin (14.18-b)

=300√36

0.88 = 43 ft (14.18-c)

Case 2: A572 Grade 65 Steel:

1. same as in case 1

2. same as in case 1

3. Required Zx = Mu

φbFy= 76(12)

0.90(65) = 15.6 in3 Select W12X14: Zx = 17.4 in3 Note that Zx is

approximated by wd9 = (14)(12)

9 = 18.7.

4. Check compact section limits λp :

λ = hctw

= 54.3λp = 640√

Fy= 640√

65= 79.4

√

λ = bf2tf

= 8.82λp = 65√

Fy= 65√

65= 8.1 < λ Not Good

In this case the controlling limit state is local buckling of the flange.Since λp < λ < λr, as above, the section is classified as non-compact.


Draft14.4 Shear Design 14–11

5. Check the strength:Since the section is non-compact, the strength is obtained by interpolation between Mp andMr.For the flanges:

λr = 141√Fy−10

= 141√65−10 = 19.0

Mn = Mp − (Mp −Mr)(λ−λpλr−λp ) ≤Mp

Mp = ZxFy = (17.4) in3(65) ksi(12) in/ft

= 94.2 k.ft

Mr = Sx(Fy − Fr) = (14.9) in3(65−10) ksi(12) in/ft

= 68.3 k.ft.

Mn = 94.2− (94.2− 68.3)(

8.8−8.119.0−8.1

)= 92.5 k.ft

φbMn = 0.90(92.5) = 83.25 k.ft > Mu√

Therefore provide W12X14 section.

14.4 Shear Design

27 Except for very short span beams, shear seldom govern. Thus sections should first be selected on thebasis of flexural reauirements, and shear should be checked next.

28 as for beams

Vd = φvVn; φv = 0.9 (14.19)

29 There are three failure modes possible:

Web yielding

Inelastic web buckling

Elastic web buckling

30 The AISC requirements for shear design are summarized below

Web Yielding htw≤ 418√

FywVn = 0.6FywAw

Inel. Web Buckling 418√Fyw

< htw≤ 523√

FywVn = 0.6FywAw

418√Fywhtw

Elast. Web Buckling 523√Fyw

< htw

Vn = 132,000Aw

( htw

)2

(14.20)and are shown in Fig. ??.

14.5 Deflections

31 Deflections of beams, under live load, are often curtailed in order not to impair serviceability (plastercracking, excessive vibration).

32 The AISC code states in Sect. L3.1 that deformations of structural members and systems due toservice loads shall not impair the serviceability of the structure. However no specific limits are given, asthose would depend on the type of structire.



50.0 60.0 70.0 80.0 90.0 100.0h/tw

10.0

15.0

20.0

25.0

30.0

35.0

40.0

Vn/

Aw [k

si]

Fy=36 ksiFy=54 ksi

Figure 14.8: AISC Requirements for Shear Design

33 In lieu of practical design experience, it is customary to limit the live-load central span deflection to

∆ ≤ L

360(14.21)

34 It should be recalled that the midspan deflection of a simply supported, uniformly loaded beam isequal to

∆ =5

384wL4

EIx(14.22)

14.6 Complete Design Example

A simply supported beam with a span of 20 ft supports a dead load of 1.0 k/ft, and a live load of 1.5k/ft in addition to its own weight (estimate dead load to 40 lbs/ft). Select the most economical W shapeusing A-36, determine the maximum distance for lateral bracing of the compression flanges, and checkfor deflection which should not exceed L/360.

WD = 1.0 + 0.04 = 1.04 k/ft (14.23-a)WL = 1.5 k/ft (14.23-b)Wu1 = 1.4wd = 1.4(1.04) = 1.456 k/ft (14.23-c)Wu2 = 1.2wd + 1.6wL = 1.2(1.04) + 1.6(1.5) = 3.648 k/ft ✛ (14.23-d)

Mu =wuL

2

8=

3.65(20)2(12)8

= 2, 189 k.in (14.23-e)

Vu =wuL

2=

(3.65)(20)2

= 36.5 k (14.23-f)

Zx =Mu

φbFy=

2, 189(0.9)(36)

= 67.6 in3 ⇒W16x40, Zx = 72.9 in3 (14.23-g)

bf2tf

= 6.9 <65√36

= 10.8√

(14.23-h)

h

tw= 46.6 <

640√36

= 106.7√

(14.23-i)

h

tw= 46.6 <

418√36

= 69.7√

(14.23-j)

φvVn = 0.9(0.6Fy) dtw︸︷︷︸Aw

= (0.9)(0.6)(36)(16.01)(0.305) = 94.9 k > 36.5√

(14.23-k)


Draft14.6 Complete Design Example 14–13

∆ =5

384wLL

4

EIx=

5384

(1.5/12) k/ in(20)4(12)4 in4

(29, 000) ksi(518) in4= 0.36 in (14.23-l)

L

360=

(20)(12)360

= 0.67 in > 0.36 in√

(14.23-m)

rmin = 1.57 in (14.23-n)

Lp =300rmin√

Fy=

(300)(1.57)√36

= 78.5 in (14.23-o)


Draft

Chapter 15

COLUMN STABILITY

15.1 Introduction; Discrete Rigid Bars

15.1.1 Single Bar System

1 Let us begin by considering a rigid bar connected to the support by a spring and axially loaded at theother end, Fig. 15.1. Taking moments about point A:

��

��

��

��

��

��

��

��

��

��

��

��

P

B

k A

B

P

∆

θ

A

θ

P

Stable Equilibrium

Neutral Equilibrium

Unstable Equilibrium

(stable)

L

Lk

sin θθ

Lk

��

��

��

��

��

��

Stable Neutral Unstable

Figure 15.1: Stability of a Rigid Bar

ΣMA = P∆− kθ = 0 (15.1-a)∆ = Lθ for small rotation (15.1-b)

PθL− kθ = 0 (15.1-c)

(P − kL

) = 0 (15.1-d)

hence we have three possibilities:

Stable Equilibrium: for P < k/L, θ = 0

Neutral Equilibrium: for P = k/L, and θ can take any value

Draft15–2 COLUMN STABILITY

Unstable equilibrium: for P > k/L, θ = 0

2 Thus we introduce a critical load defined by

Pcr =k

L(15.2)

3 For large rotation, we would have

ΣMA = P∆− kθ = 0 (15.3-a)∆ = L sin(θ)PθL − kθ = 0 (15.3-b)

Pcr =k

L

θ

sin(θ)(15.3-c)

4 Because there is more than one possible path (θ) when P = Pcr, we call this point a biffurcationpoint.

5 If we now assume that there is an initial imperfection in the column, i.e. the column is initially“crooked”, Fig. 15.2, then

ΣMA = PLθ − k(θ − θ0) = 0 (15.4-a)

Pcr =k

L(1− θ0

θ) (15.4-b)

θ0

θ0

θ0

��

��

��

��

L

B

P

∆

θ

Aθ

P

Perfect System

L(1-

θ)

Lk k

k

Figure 15.2: Stability of a Rigid Bar with Initial Imperfection

15.1.2 Two Bars System

6 Next we consider the two rigid bar problem illustrated in Fig. 15.3.

ΣMB = PLθ2 − k(θ2 − θ1) = 0 (15.5-a)

⇒ −θ1 + θ2 =PL

kθ2 (15.5-b)

ΣMA = PLθ1 + k(θ2 − θ1)− kθ1 = 0 (15.5-c)

⇒ 2θ1 − θ2 =PL

kθ1 (15.5-d)


Draft15.1 Introduction; Discrete Rigid Bars 15–3

θ2

θ1

k( - ) θ2

θ1

k( - )

θ1

θ2

θ1

��

��

��

��

��

��

1

1.618

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

L

k

L

k

A

B

C

P

C

A

B

P

P

θ2

B

k

θ1

LB

C

1

.618

P

L

P

P

P P

Figure 15.3: Stability of a Two Rigid Bars System

7 Those two equations can be cast in matrix form[2 −1−1 1

]{θ1θ2

}= λ

{θ1θ2

}(15.6)

where λ = PL/k, this is an eigenvalue formulation and can be rewritten as[2− λ −1−1 1− λ

]{θ1θ2

}= λ

{00

}(15.7)

This is a homogeneous system of equation, and it can have a non-zero solution only if its determinant isequal to zero ∣∣∣∣ 2− λ −1

−1 1− λ∣∣∣∣ = 0 (15.8-a)

2− λ− 2λ+ λ2 − 1 = 0 (15.8-b)λ2 − 3λ+ 1 = 0 (15.8-c)

λ1,2 =3±√9− 4

2=

3±√52

(15.8-d)

8 Hence we now have two critical loads:

Pcr1 =3−√5

2k

L= 0.382

k

L(15.9)

Pcr2 =3 +√

52

k

L= 2.618

k

L(15.10)

9 We now seek to determine the deformed shape for each of the first critical loads. It should be notedthat since the column will be failing at the critical buckling load, we can not determine the absolutevalues of the deformations, but rather the shape of the buckling column.

λ1 =3−√5

2(15.11-a)[

2− 3−√5

2 −1−1 1− 3−√

52

]{θ1θ2

}=

{00

}(15.11-b)



[1+

√5

2 −1−1 1− −1+√

52

]{θ1θ2

}=

{00

}(15.11-c)

[1.618 −1−1 0.618

]{θ1θ2

}=

{00

}(15.11-d)

we now arbitrarily set θ1 = 1, then θ2 = 1/0.618 = 1.618, thus the first eigenmode is

{θ1θ2

}={

11.618

}(15.12)

10 Note that we can determine the deformed shape upon buckling but not the geometry.

11 Finally, we examine the second mode shape loads

λ2 =3 +√

52

(15.13-a)[2− 3+

√5

2 −1−1 1− 3+

√5

2

]{θ1θ2

}=

{00

}(15.13-b)

[1−√

52 −1−1 1− −1−√

52

]{θ1θ2

}=

{00

}(15.13-c)

[ −0.618 −1−1 −1.618

]{θ1θ2

}=

{00

}(15.13-d)

we now arbitrarily set θ1 = 1, then θ2 = −1/1.618 = −0.618, thus the second eigenmode is

{θ1θ2

}={

1−0.618

}(15.14)

15.1.3 ‡Analogy with Free Vibration

12 The problem just considered bears great ressemblence with the vibration of a two degree of freedommass spring system, Fig. 15.4.

m = m1

2m

= 2

m

1k = k

u2

1u

1u

u2

u2

k u 2

k(u -u ) 2 1

2m u 2

..

m u

1

..

k u

1

��

��

��

��

k = k2

k = k3

Figure 15.4: Two DOF Dynamic System

13 Each mass is subjected to an inertial force equals to the mass times the acceleration, and the springforce:

2mu2 + ku2 + k(u2 − u1) = 0 (15.15-a)


Draft15.2 Continuous Linear Elastic Systems 15–5

mu1 + ku1 + k9u2 − u10 = 0 (15.15-b)

or in matrix form [m 00 2m

]︸︷︷︸

M

{u1u2

}︸︷︷︸

U

+[

2k −k−k 2k

]︸︷︷︸

K

{u1u2

}︸︷︷︸

U

(15.16)

14 The characteristic equation is |K− λM| where λ = ω2, and ω is the natural frequency.∣∣∣∣ 2h− λ −h−h 2h− 2λ

∣∣∣∣ = 0 (15.17)

where h = k/m. This reduces to the polynomial

λ2 − 3hλ+32h2 = 0 (15.18)

Solving, λ = (3±−√3)h/2 or

ω1 = 0.796√k/m (15.19-a)

ω1 = 1.538√k/m (15.19-b)

15 To find the mode shapes φ1 and φ2 (relative magnitudes of the DOF) we substitute in the characteristicequation and set the first element equal to 1:

φ1 ={

1.0001.3660

}and φ1 =

{1.000−0.3660

}(15.20)

15.2 Continuous Linear Elastic Systems

16 Column buckling theory originated with Leonhard Euler in 1744.

17 An initially straight member is concentrically loaded, and all fibers remain elastic until buckling occur.

18 For buckling to occur, it must be assumed that the column is slightly bent as shown in Fig. 15.5.Note, in reality no column is either perfectly straight, and in all cases a minor imperfection is present.

Px

x

L

x and y are

principal axes

Slightly bent position

y, v

Figure 15.5: Euler Column

19 Two sets of solutions will be presented, in both cases the equation of equilibrium is written forthe deformed element.



15.2.1 Lower Order Differential Equation

20 At any location x along the column, the imperfection in the column compounded by the concentricload P , gives rise to a moment

Mz = Pv (15.21)

Note that the value of y is irrelevant.

21 Recalling thatd2v

dx2=Mz

EI(15.22)

upon substitution, we obtain the following differential equation

d2v

dx2− P

EIv = 0 (15.23)

22 Letting k2 = PEI , the solution to this second-order linear differential equation is

v = −A sinkx−B cos kx (15.24)

23 The two constants are determined by applying the boundary conditions

1. v = 0 at x = 0, thus B = 0

2. v = 0 at x = L, thusA sinkL = 0 (15.25)

24 This last equation can be satisfied if: 1) A = 0, that is there is no deflection; 2) kL = 0, that is noapplied load; or 3)

kL = nπ (15.26)

Thus buckling will occur if PcrEI =

(nπL

)2 or

Pcr =n2π2EI

L2

25 The fundamental buckling mode, i.e. a single curvature deflection, will occur for n = 1; Thus Eulercritical load for a pinned column is

Pcr =π2EI

L2(15.27)

26 The corresponding critical stress is

σcr =π2E(L

rmin

)2 (15.28)

where I = Ar2min.

27 Note that buckling will take place with respect to the weakest of the two axis.

15.2.2 Higher Order Differential Equation

15.2.2.1 Derivation

28 In the preceding approach, the buckling loads were obtained for a column with specified boundaryconditons. A second order differential equation, valid specifically for the member being analyzed wasused.



29 In the next approach, we derive a single fourth order equation which will be applicable to any columnregardelss of the boundary conditions.

30 Considering a beam-column subjected to axial and shear forces as well as a moment, Fig. 15.6 (noteanalogy with cable structure, Fig. 6.1), taking the moment about i for the beam segment and assuming

i

i

jδv

δ

δV

M

δx

δ

δx

x

w

P

P

V+ dx

M+ dxθi

θj

w(x)

P P

y, u dx

x

w

M

P

Pdx

��

��

Figure 15.6: Simply Supported Beam Column; Differential Segment; Effect of Axial Force P

the angle dvdx between the axis of the beam and the horizontal axis is small, leads to

M −(M +

dM

dxdx

)+ w

(dx)2

2+(V +

dV

dxdx

)︸︷︷︸−P

(dv

dx

)dx︸︷︷︸ = 0 (15.29)

Note that the first underbraced term is identical to the one used in earlier derivation of the beam’sdifferential equation.

31 neglecting the terms in dx2 which are small, and then differentiating each term with respect to x, weobtain

d2M

dx2− dVdx− P d

2v

dx2= 0 (15.30)

32 However, considering equilibrium in the y direction gives

dV

dx= −w (15.31)

33 From beam theory, neglecting axial and shear deformations, we have

M = −EI d2v

dx2(15.32)

34 Substituting Eq. 15.31 and 15.32 into 15.30, and assuming a beam of uniform cross section, we obtain

EId4v

dx4− P d

2v

dx2= w (15.33)

substituting λ =√

PEI , and w = 0 we finally obtain

d4v

dx4+ λ2

d2v

dx2= 0 (15.34)

35 Again, we note that by considering equilibrium in the deformed state we have introduced the secondterm to what would otherwise be the governing differential equation for flexural members (beams).Finally, we note the analogy between this equation, and the governing differential equation for a cablestructure, Eq. 6.12.

36 The general solution of this fourth order differential equation to any set of boundary conditions is

v = C1 + C2x+ C3 sinλx+ C4 cosλx (15.35)



Essential Natural(Dirichlet) (Neumann)

vd3v

dx3(V )

dv

dx

d2v

dx2(M)

Table 15.1: Essential and Natural Boundary Conditions for Columns

37 The constants C′s are obtained from the boundary condtions. For columns, those are shown in Table?? The essential boundary conditions are associated with displacement, and slope, the natural ones withshear and moments (through their respective relationships with the displacement).

38 We note that at each node, we should have two boundary conditions, all combinations are possibleexcept pairs from the same raw (i.e. we can not have known displacement and shear, or known slopeand moment).

15.2.2.2 Hinged-Hinged Column

39 If we consider again the stability of a hinged-hinged column, the boundary conditions are displacement(v) and moment ( d

2vdx2EI) equal to zero at both ends1, or

v = 0, d2vdx2 = 0 at x = 0

v = 0, d2vdx2 = 0 at x = L

(15.36)

substitution of the two conditions at x = 0 leads to C1 = C4 = 0. From the remaining conditions, weobtain

C3 sinλL + C2L = 0 (15.37-a)−C3k2 sinλL = 0 (15.37-b)

these relations are satisfied either if C2 = C3 = 0 or if sinλL = C2 = 0. The first alternative leads tothe trivial solution of equilibrium at all loads, and the second to λL = nπ for n = 1, 2, 3 · · ·. The criticalload is

Pcr =n2π2EI

L2(15.38)

which was derived earlier using the lower order differential equation.

40 The shape of the buckled column isy = C3 sin

nπx

L(15.39)

and only the shape (but not the geometry) can be determined. In general, we will assume C3 = 1 andplot y.

15.2.2.3 Fixed-Fixed Column

41 We now consider a column which is restrained against rotation at both ends, the boundary conditionsare given by:

v(0) = 0 = C1 + C4 (15.40-a)v′(0) = 0 = C2 + C3λ (15.40-b)v(L) = 0 = C1 + C2L+ C3 sinλL + C4 cosλL (15.40-c)v′(L) = 0 = C2 + C3λ cos(λL)− C4 sin(λL) (15.40-d)

1It will be shown in subsequent courses, that the former BC is an essential B.C., and the later a natural B.C.



those equations can be set in matrix form

1 0 0 11 1 λ 01 L sinλL cosλL0 1 λ cosλL −λ sinλL

C1C2C3C4

=

0000

(15.41-a)

∣∣∣∣∣∣∣∣1 0 0 11 1 λ 01 L sinλL cosλL0 1 λ cosλL −λ sinλL

∣∣∣∣∣∣∣∣ = 0 (15.41-b)

The determinant is obtain from ∣∣∣∣∣∣1 λ 0L sinλL cosλL1 λ cosλL −λ sinλL

∣∣∣∣∣∣−∣∣∣∣∣∣

1 1 λ1 L sinλL0 1 λ cosλL

∣∣∣∣∣∣ = 0 (15.42-a)

∣∣∣∣ sinλL cosλLλ cosλL −λ sinλL

∣∣∣∣− λ∣∣∣∣ L cosλL

1 −λ sinλL

∣∣∣∣+∣∣∣∣ 1 sinλL

0 λ cosλL

∣∣∣∣− λ∣∣∣∣ 1 L

0 1

∣∣∣∣ = 0 (15.42-b)

−2λ+ λ2L sinλL+ 2λ cosλL = 0 (15.42-c)

The first solution, λ = 0 is a trivial one, and the next one

λL sinλL + 2 cosλL− 2 = 0 (15.43-a)λL sinλL = 2(1− cosλL) (15.43-b)

The solution to this transcendental equation is

λL = 2nπ (15.44-a)√P

EI=

2nπL

(15.44-b)

thus the critical load and stresses are given by

Pcr =4n2π2EIL2

(15.45)

σcr =4n2π2E(L/r)2

(15.46)

42 The deflected shape (or eigenmodes) can be obtained by substituting the value of λ into the c′s.

15.2.2.4 Fixed-Hinged Column

43 Next we consider a column with one end fixed (at x = L), and one end hinged (at x = 0). Theboundary conditions are

v = 0, d2vdx2 = 0 at x = 0

v = 0, dvdx = 0 at x = L

(15.47)

The first two B.C. yield C1 = C4 = 0, and the other two

sinλL − λL cosλL = 0 (15.48)

But since cosλL can not possibly be equal to zero, the preceding equation can be reduced to

tanλL = λL (15.49)

which is a transcendental algebraic equation and can only be solved numerically. We are essentiallylooking at the intersection of y = x and y = tanx, Fig. 15.7 and the smallest positive root is λL = 4.4934,since k2 = P

EI , the smallest critical load is

Pcr =(4.4934)2

L2EI =

π2

(0.699L)2EI (15.50)

Note that if we were to solve for x such that v,xx = 0 (i.e. an inflection point), then x = 0.699L.



0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0−10.0

−8.0

−6.0

−4.0

−2.0

0.0

2.0

4.0

6.0

8.0

10.0

Figure 15.7: Solution of the Tanscendental Equation for the Buckling Load of a Fixed-Hinged Column

15.2.3 Effective Length Factors K

44 Recall that the Euler buckling load was derived for a pinned column. In many cases, a column willhave different boundary conditions. It can be shown that in all cases, the buckling load would be givenby

Pcr =π2EI

(KL)2(15.51)

where K is called effective length factor, and KL is the effective length. and

σcr =π2E(KLrmin

)2 (15.52)

The ratio KLrmin

is termed the slenderness ratio. Note that rmin should be the smallest radius ofgyration in the unbraced direction(s).

45 The effective length, can only be determined by numerical or approximate methods, and is the distancebetween two adjacent (real or virtual) inflection points, Fig. 15.8, 15.9

46 The most widely used charts for the effective length determination are those produced by the Struc-tural Stability Research Council. The alignment chart, for an individual column, Fig. 15.10 is shown inFig. 15.11. It should be noted that this chart assumes that all members are still in the elastic range.

47 The use of the alignment chart involces computing G at each end of the column using the followingformula

Ga =

∑ IcLc∑ IgLg

(15.53)

where Ga is the stiffness at end a of the column, Ic, Ig are the moment of inertias of the columns andgirders respectively. The summation must include only those members which are rigidly connected tothat joint and lying in the plane for which buckling is being considered.



P

P PP

P

PP

P

KL = KL =

KL =

KL <L12

LL L L

L0.7

LL

(a) End rotationsunrestrained

(b) End rotations fully restrained

(c) One endrestrained, otherunrestrained

(d) Partially restrainedat each end

Figure 15.8: Column Effective Lengths

P P P P

P P P P

L L

LL

0.7L<KL<L

KL2

KL>2L

0.5L<KL<0.7L

L<KL<2L

∆

∆

(b) Unbraced frame, hinged base(a) Braced frame, hinged base

(c) Braced frame, fixed base (d) Unbraced frame, fixed base

Figure 15.9: Frame Effective Lengths



Figure 15.10: Column Effective Length in a Frame

48 Hence, once Ga and Gb are determined, those values are connected by a straight line in the appropriatechart, and k is the point of intersection of that line with the midle axis.

49 Alternatively :-)

GAGB4

( πK

)2+(GA +GB

2

)(1− π/K

tanπ/K

)+

2 tanπ/2Kπ/K

= 1 (15.54)

(Ref. McGuire P. 467).

15.3 Inelastic Columns

50 There are two limiting loads for a column

1. Yielding of the gross section Pcr = FyAg, which occirs in short stiff columns

2. Elastic (Euler) buckling, Eq. 15.51 Pcr = π2EI(KL)2

, in long slender columns.

51 Those two expression are asymptotic values for actual column buckling. Intermediary failure loadsare caused by the presence of residual stresses which in turn give rise to inelastic buckling.

52 Inelastic buckling buckling occurs when the stresses (average) have not yet reached the yield stress,and is based on the tangent modulus Et which is lower than the initial modulus E.

53 Residual stresses (caused by uneven cooling) will initiate inelastic buckling, Fig. 15.12.

54 The inelastic buckling curve has to be asymptotic to both the Euler elastic buckling (for slendercolumns), and to the yield stress (for stiff columns).

55 The Structural Stability Research Council (SSRC) has proposed a parabolic curve which provides atransition between elastic buckling and yielding, thus accounting for the presence of residual stresses


Draft15.3 Inelastic Columns 15–13

Sidesway Inhibited

Ga

0.

0.1

0.2

0.3

0.4

0.5

0.60.70.8

1.

2.

3.

5.

10.50.∞

K

0.5

0.6

0.7

0.8

0.9

1.0Gb

0.

0.1

0.2

0.3

0.4

0.5

0.60.70.8

1.

2.

3.

5.

10.50.∞

Sidesway Uninhibited

Ga

0

1.

2.

3.

4.

5.

6.7.8.9.10.

20.30.

50.100.

∞K

1.

1.5

2.

3.

4.

5.

10.20.∞

Gb

0

1.

2.

3.

4.

5.

6.7.8.9.10.

20.30.

50.100.

∞

Figure 15.11: Standard Alignment Chart (AISC)

ydσ

ydσ

kLr

kLr

2σ = π E

( )2

E T < E

ε

σ

rσ

σ Euler Buckling (Linear Elastic)

Gross Yielding

InelasticBuckling

Proportional Limit Proportional Limit

E

Effect of Residual Stresses

Figure 15.12: Inelastic Buckling



and the resulting inelastic buckling, Fig. 15.13.

σcr = σy

[1− σy

4π2E

(KL

rmin

)2](15.55)

0 20 40 60 80 100 120 140 160 180 200Slenderness Ratio KL/rmin

0

10

20

30

40

50S

tres

s σ

[ksi

]Fy=36 ksi; E=29,000 ksi

EulerSSRC

Figure 15.13: Euler Buckling, and SSRC Column Curve


Draft

Chapter 16

STEEL COMPRESSIONMEMBERS

1 This chapter will cover the elementary analysis and design of steel column according to the LRFDprovisions.

16.1 AISC Equations

2 By introducing a slenderness parameter λc (not to be confused with the slenderness ratio) definedas

λ2c =Fy

FEulercr

=Fyπ2E(KLrmin

)2 (16.1-a)

λc =KL

rmin

√Fyπ2E

(16.1-b)

Hence, this parameter accounts for both the slenderness ratio as well as steel material properties.This new parameter is a more suitable one than the slenderness ratio (which was a delimeter for elasticbuckling) for the inelastic buckling.

3 Equation 15.55 becomesFcrFy

= 1− λ2c

4for λc ≤

√2 (16.2)

4 When λc >√

2, then Euler equation applies

FcrFy

=1λ2c

for λc ≥√

2 (16.3)

Hence the first equation is based on inelastic buckling (with gross yielding as a limiting case), and thesecond one on elastic buckling. The two curves are tangent to each other.

5 The above equations are valid for concentrically straight members.

6 The previous two equations, were modified to account for an initial out-of-straightness of about 1/1500.This will result in, Fig. 16.1:

Fcr = 0.658λ2cFy for λc ≤ 1.5 (16.4)

Fcr =0.877λ2c

Fy for λc > 1.5 (16.5)

Draft16–2 STEEL COMPRESSION MEMBERS

where

λc =KL

πr

√FyE

(16.6)

0 20 40 60 80 100 120 140 160 180 200Slenderness Ratio KL/rmin

0

10

20

30

40

50

Str

ess

σ [k

si]

Fy=36 ksi; E=29,000 ksi

SSRC0.6584

λ2c; λc<1.5

0.877λc

2Fy; λc>1.5

λ >1.5

λc

<1.5

c

Inelastic Buckling

Elastic Buckling

Figure 16.1: SSRC Column Curve and AISC Critical Stresses

7 Fcr can readily be determined from Figure 16.2 or from Table 16.1 and 16.2.

8 The AISC code introduces a reduction factorQ into Eq. 16.7-a and 16.7-b when certan width/thicknesslimitations (LRFD Table B5.1) are not satisfied, and local buckling may occur prior to overall buckling.

Fcr = 0.658Qλ2cFy for λc

√Q ≤ 1.5 (16.7-a)

Fcr =0.877λ2c

Fy for λc√Q > 1.5 (16.7-b)

since Q = 1 for all rolled H sections (standard W, S, and M shapes) it will be neglected in this course.

9 Finally, the AISC introduces a stiffness reduction factor (SRF) for columns when the nomographs areused. Such a reduction would account for the discrepancy between elastic buckling (assumed in thenomographs) and inelastic buckling which occur for relatively high Pu/A values, Table 16.3.

16.2 LRFD Equations

10 The strength requirement for columns in load and resistance factor design is stated as

φcPn ≥ Pu (16.8)

where:φc strength reduction factor, 0.85Pn nominal strength=AgFcrFcr Eq. 16.7-a and 16.7-bPu factored service load.


Draft16.2 LRFD Equations 16–3

Klrmin

λc φcFcrKl

rminλc φcFcr

Klrmin

λc φcFcrKl

rminλc φcFcr

Klrmin

λc φcFcr

1 0.01 30.60 41 0.46 28.01 81 0.91 21.66 121 1.36 14.16 161 1.81 8.232 0.02 30.59 42 0.47 27.89 82 0.92 21.48 122 1.37 13.98 162 1.82 8.133 0.03 30.59 43 0.48 27.76 83 0.93 21.29 123 1.38 13.80 163 1.83 8.034 0.04 30.57 44 0.49 27.63 84 0.94 21.11 124 1.39 13.62 164 1.84 7.935 0.06 30.56 45 0.50 27.51 85 0.95 20.92 125 1.40 13.44 165 1.85 7.846 0.07 30.54 46 0.52 27.37 86 0.96 20.73 126 1.41 13.27 166 1.86 7.747 0.08 30.52 47 0.53 27.24 87 0.98 20.54 127 1.42 13.09 167 1.87 7.658 0.09 30.50 48 0.54 27.10 88 0.99 20.35 128 1.44 12.92 168 1.88 7.569 0.10 30.47 49 0.55 26.97 89 1.00 20.17 129 1.45 12.74 169 1.90 7.4710 0.11 30.44 50 0.56 26.83 90 1.01 19.98 130 1.46 12.57 170 1.91 7.3811 0.12 30.41 51 0.57 26.68 91 1.02 19.79 131 1.47 12.40 171 1.92 7.3012 0.13 30.37 52 0.58 26.54 92 1.03 19.60 132 1.48 12.23 172 1.93 7.2113 0.15 30.33 53 0.59 26.39 93 1.04 19.41 133 1.49 12.06 173 1.94 7.1314 0.16 30.29 54 0.61 26.25 94 1.05 19.22 134 1.50 11.88 174 1.95 7.0515 0.17 30.24 55 0.62 26.10 95 1.07 19.03 135 1.51 11.71 175 1.96 6.9716 0.18 30.19 56 0.63 25.94 96 1.08 18.84 136 1.53 11.54 176 1.97 6.8917 0.19 30.14 57 0.64 25.79 97 1.09 18.65 137 1.54 11.37 177 1.99 6.8118 0.20 30.08 58 0.65 25.63 98 1.10 18.46 138 1.55 11.20 178 2.00 6.7319 0.21 30.02 59 0.66 25.48 99 1.11 18.27 139 1.56 11.04 179 2.01 6.6620 0.22 29.96 60 0.67 25.32 100 1.12 18.08 140 1.57 10.89 180 2.02 6.5921 0.24 29.90 61 0.68 25.16 101 1.13 17.89 141 1.58 10.73 181 2.03 6.5122 0.25 29.83 62 0.70 24.99 102 1.14 17.70 142 1.59 10.58 182 2.04 6.4423 0.26 29.76 63 0.71 24.83 103 1.16 17.51 143 1.60 10.43 183 2.05 6.3724 0.27 29.69 64 0.72 24.66 104 1.17 17.32 144 1.61 10.29 184 2.06 6.3025 0.28 29.61 65 0.73 24.50 105 1.18 17.13 145 1.63 10.15 185 2.07 6.2326 0.29 29.53 66 0.74 24.33 106 1.19 16.94 146 1.64 10.01 186 2.09 6.1727 0.30 29.45 67 0.75 24.16 107 1.20 16.75 147 1.65 9.87 187 2.10 6.1028 0.31 29.36 68 0.76 23.99 108 1.21 16.56 148 1.66 9.74 188 2.11 6.0429 0.33 29.27 69 0.77 23.82 109 1.22 16.37 149 1.67 9.61 189 2.12 5.9730 0.34 29.18 70 0.79 23.64 110 1.23 16.18 150 1.68 9.48 190 2.13 5.9131 0.35 29.09 71 0.80 23.47 111 1.24 16.00 151 1.69 9.36 191 2.14 5.8532 0.36 28.99 72 0.81 23.29 112 1.26 15.81 152 1.70 9.23 192 2.15 5.7933 0.37 28.90 73 0.82 23.11 113 1.27 15.62 153 1.72 9.11 193 2.16 5.7334 0.38 28.79 74 0.83 22.94 114 1.28 15.44 154 1.73 9.00 194 2.18 5.6735 0.39 28.69 75 0.84 22.76 115 1.29 15.25 155 1.74 8.88 195 2.19 5.6136 0.40 28.58 76 0.85 22.58 116 1.30 15.07 156 1.75 8.77 196 2.20 5.5537 0.41 28.47 77 0.86 22.40 117 1.31 14.88 157 1.76 8.66 197 2.21 5.5038 0.43 28.36 78 0.87 22.21 118 1.32 14.70 158 1.77 8.55 198 2.22 5.4439 0.44 28.25 79 0.89 22.03 119 1.33 14.52 159 1.78 8.44 199 2.23 5.3940 0.45 28.13 80 0.90 21.85 120 1.35 14.34 160 1.79 8.33 200 2.24 5.33

Table 16.1: Design Stress φFcr for Fy=36 ksi in Terms of KLrmin



Klrmin

λc φcFcrKl

rminλc φcFcr

Klrmin

λc φcFcrKl

rminλc φcFcr

Klrmin

λc φcFcr

1 0.01 42.50 41 0.54 37.58 81 1.07 26.31 121 1.60 14.57 161 2.13 8.232 0.03 42.49 42 0.56 37.36 82 25.99 122 1.61 14.33 162 2.14 8.133 0.04 42.47 43 0.57 37.13 83 1.10 25.68 123 1.63 14.10 163 2.15 8.034 0.05 42.45 44 0.58 36.89 84 1.11 25.37 124 1.64 13.88 164 2.17 7.935 0.07 42.42 45 0.59 36.65 85 1.12 25.06 125 1.65 13.66 165 2.18 7.846 0.08 42.39 46 0.61 36.41 86 1.14 24.75 126 1.67 13.44 166 2.19 7.747 0.09 42.35 47 0.62 36.16 87 1.15 24.44 127 1.68 13.23 167 2.21 7.658 0.11 42.30 48 0.63 35.91 88 1.16 24.13 128 1.69 13.02 168 2.22 7.569 0.12 42.25 49 0.65 35.66 89 1.18 23.82 129 1.71 12.82 169 2.23 7.4710 0.13 42.19 50 0.66 35.40 90 1.19 23.51 130 1.72 12.62 170 2.25 7.3811 0.15 42.13 51 0.67 35.14 91 1.20 23.20 131 1.73 12.43 171 2.26 7.3012 0.16 42.05 52 0.69 34.88 92 1.22 22.89 132 1.74 12.25 172 2.27 7.2113 0.17 41.98 53 0.70 34.61 93 1.23 22.58 133 1.76 12.06 173 2.29 7.1314 0.19 41.90 54 0.71 34.34 94 1.24 22.27 134 1.77 11.88 174 2.30 7.0515 0.20 41.81 55 0.73 34.07 95 1.26 21.97 135 1.78 11.71 175 2.31 6.9716 0.21 41.71 56 0.74 33.79 96 1.27 21.66 136 1.80 11.54 176 2.33 6.8917 0.22 41.61 57 0.75 33.51 97 1.28 21.36 137 1.81 11.37 177 2.34 6.8118 0.24 41.51 58 0.77 33.23 98 1.30 21.06 138 1.82 11.20 178 2.35 6.7319 0.25 41.39 59 0.78 32.95 99 1.31 20.76 139 1.84 11.04 179 2.37 6.6620 0.26 41.28 60 0.79 32.66 100 1.32 20.46 140 1.85 10.89 180 2.38 6.5921 0.28 41.15 61 0.81 32.38 101 1.33 20.16 141 1.86 10.73 181 2.39 6.5122 0.29 41.02 62 0.82 32.09 102 1.35 19.86 142 1.88 10.58 182 2.41 6.4423 0.30 40.89 63 0.83 31.79 103 1.36 19.57 143 1.89 10.43 183 2.42 6.3724 0.32 40.75 64 0.85 31.50 104 1.37 19.27 144 1.90 10.29 184 2.43 6.3025 0.33 40.60 65 0.86 31.21 105 1.39 18.98 145 1.92 10.15 185 2.45 6.2326 0.34 40.45 66 0.87 30.91 106 1.40 18.69 146 1.93 10.01 186 2.46 6.1727 0.36 40.29 67 0.89 30.61 107 1.41 18.40 147 1.94 9.87 187 2.47 6.1028 0.37 40.13 68 0.90 30.31 108 1.43 18.11 148 1.96 9.74 188 2.48 6.0429 0.38 39.97 69 0.91 30.01 109 1.44 17.83 149 1.97 9.61 189 2.50 5.9730 0.40 39.79 70 0.93 29.70 110 1.45 17.55 150 1.98 9.48 190 2.51 5.9131 0.41 39.62 71 0.94 29.40 111 1.47 17.26 151 2.00 9.36 191 2.52 5.8532 0.42 39.43 72 0.95 29.09 112 1.48 16.98 152 2.01 9.23 192 2.54 5.7933 0.44 39.25 73 0.96 28.79 113 1.49 16.71 153 2.02 9.11 193 2.55 5.7334 0.45 39.06 74 0.98 28.48 114 1.51 16.42 154 2.04 9.00 194 2.56 5.6735 0.46 38.86 75 0.99 28.17 115 1.52 16.13 155 2.05 8.88 195 2.58 5.6136 0.48 38.66 76 1.00 27.86 116 1.53 15.86 156 2.06 8.77 196 2.59 5.5537 0.49 38.45 77 1.02 27.55 117 1.55 15.59 157 2.08 8.66 197 2.60 5.5038 0.50 38.24 78 1.03 27.24 118 1.56 15.32 158 2.09 8.55 198 2.62 5.4439 0.52 38.03 79 1.04 26.93 119 1.57 15.07 159 2.10 8.44 199 2.63 5.3940 0.53 37.81 80 1.06 26.62 120 1.59 14.82 160 2.11 8.33 200 2.64 5.33

Table 16.2: Design Stress φFcr for Fy=50 ksi in Terms of KLrmin


Draft16.2 LRFD Equations 16–5

0 20 40 60 80 100 120 140 160 180 200Slenderness ratio, KL/r

0

10

20

30

40

50

60

70

80

φ F

cr (

ksi)

Fy=36 ksiFy=50 ksiFy=60 ksiFy=70 ksiFy=100 ksi

Figure 16.2: Fcr versus KL/r According to LRFD, for Various Fy

Pu/A Fy Pu/A Fy36 ksi 50 ksi 36 ksi 50 ksi

42 0.03 26 0.38 0.8241 0.09 25 0.45 0.8540 0.16 24 0.52 0.8839 0.21 23 0.58 0.9038 0.27 22 0.65 0.9337 0.33 21 0.70 0.9536 0.38 20 0.76 0.9735 0.44 19 0.81 0.9834 0.49 18 0.85 0.9933 0.53 17 0.89 1.0032 0.58 16 0.92 1.0031 0.63 15 0.85 1.0030 0.05 0.67 14 0.97 1.0029 0.14 0.71 13 0.99 1.0028 0.22 0.75 12 1.00 1.0027 0.30 0.79 11 1.00 1.00

Table 16.3: Stiffness Reduction Factors (AISC).



16.3 Examples

16.3.1 Evaluation

11 In this class of problems, we seek to assess the load carrying capacity of a given column. This, incontrast to design problems, is a straightforward application of the governing equations.

Example 16-1: Unbraced Column Evaluation, (?)

Calculate the maximum factored load Pu that a 20 ft. long, unbraced A36 W 14x53 column canhold. (K = 1.0)Solution:

1. The W14x53 section has, (Sect. 3.6) A = 15.6 in2, rx = 5.89 in, and ry = 1.92 in.2. Since the column is not braced, the weak axis will control

KL

ry=

(1.0)(20) ft(12) in/ft

(1.92) in= 125 dimensionless

3. The slenderness parameter

λc =KL

πr

√FyE

=125π

√36, 000

29× 106= 1.401 < 1.5

4. Since λc = 1.401 < 1.5, inelastic behavior controls and we use Eq. 16.7-a

Fcr = 0.658λ2cFy = 0.6581.401

2(36) ksi = 15.81 ksi

5. The design capacity of the section is thus

Pu = φcPn = φcAgFcr = (0.85)(15.6) in2(15.81) ksi = 209.6 k

Example 16-2: Braced Column Evaluation, (?)

Calculate the maximum load, Pu for a 20 ft. long, A36 W24x104 column with K = 0.8. Thenrecalculate if the column is braced at midheight in the weak axis (still assume K = 0.8).Solution:

Unbraced Case:

1. The W24x104 section has, (Sect. 3.6) A = 30.6 in2, rx = 10.1 in, and ry = 2.91 in.

Since the column is not braced, the weak axis will control

KL

ry=

(0.8)(20) ft(12) in/ft

(2.91) in= 65.98


λc =KL

πr

√FyE

=65.98π

√36, 000

29× 106= .7399 < 1.5

3. Since λc = 1.401 < 1.5 inelastic behavior controls and we use Eq. 16.7-a

Fcr = 0.658λ2cFy = 0.658.7399

2(36) ksi = 28.62 ksi





Braced Case:

5. We compute the slenderness ratios for both axis

KLrx

= (0.8)(20) ft(12) in/ft(10.1) in = 19.01

KLry

= (0.8)(10) ft(12) in/ft(2.91) in = 32.99 ✛


λc =KL

πr

√FyE

=32.99π

√36, 000

29× 106= .369 < 1.5


Fcr = 0.658λ2cFy = 0.658.369

2(36) ksi = 34 ksi


Pu = φcPn = φcAgFcr = (0.85)(30.6) in2(34) ksi = 884.4 k

9. We observe that bracing has increased the column capacity by 888.4−744.4744.4 = 18.8%

Example 16-3: Braced Column Evaluation, (?)

Calculate the maximum load, Pu for a 20 ft. long, A36 W12x79 column. The column is braced alongits weak axis at five feet from the top and the bottom. The column is pinned at both ends.Solution:

1. The W12x79 section has, (Sect. 3.6) A = 23.2 in2, rx = 5.34 in, and ry = 3.05 in.2. The unbraced lengths along the weak axis are 5 ft. at the ends and 10 ft. in the middle, however theunbraced length along the strong axis is 20 ft.3. We compute the slenderness ratios for both axis

KLrx

= (1.0)(20) ft(12) in/ft(5.34) in = 44.9 ✛

KLry

= (1.0)(10) ft(12) in/ft(3.05) in = 39.3


λc =KL

πr

√FyE

=44.9π

√36, 000

29× 106= .503 < 1.5


Fcr = 0.658λ2cFy = 0.658.503

2(36) ksi = 32.4 ksi





16.3.2 Design

12 Design problems are inherently more difficult than verification ones, and a trial and error proceduremust be followed.

1. Assume a design a design stress, φcFcr, typically around 25-30 ksi

2. Using the assumed φcFcr, determine Atrialg = PuφcFcr

3. Select a trial member on the basis of Atrialg

4. Determine Fcr of trial member, and compare φcPn with Pu.

5. If

(a) φcPn ≈ Pu, selection is appropriate(b) φcPn > Pu, select a smaller section(c) φcPn < Pu, select a larger section

13 Note that often the depth of the member is specified.

Example 16-4: Column Design, (?)

Find the most economical W12 section to hold a compressive load of 120 kips. Steel is A36, K = 1.0,L = 16 feet. Live load to dead load ratio is 1.0 and the column is unbraced.Solution:

1. Since the dead load and live loads are evenly split, from Sect. 7.3.3

Pu = 1.6(60) + 1.2(60) = 168 k

2. The effective length KL=16 feet, we assume a design stress of φcFcr = 25 ksi and

Atrialg =PuφcFcr

=16825

= 6.72 in2

3. Select W12x26 as a trial member (A = 7.75 in2)

KLry

= (1.0)(16) ft(12) in/ft(1.51) in = 127.1

λc = KLπr

√FyE

= 127.1π

√36,00029×106 = 1.42 < 1.5

Fcr = 0.658λ2cFy

= 0.6581.422(36) ksi = 15.4 ksi

φcPn = φcAgFcr= (0.85)(7.65) in2(15.4) ksi = 100.1 k

4. Since the design capacity of this section is less than the required one, we need a bigger section, tryW12x40, (A = 11.8 in2).

KLry

= (1.0)(16) ft(12) in/ft(1.93) in = 99.5

λc = 99.5π

√36,00029×106 = 1.116 < 1.5

Fcr = 0.6581.1162(36) ksi = 21.37 ksi

φcPn = φcAgFcr= (0.85)(11.8) in2(21.37) ksi = 214.3 k

5. Since W 12x40 has a design capacity greater than the required one, it is a satisfactory design. Becausethere is a sizeable gap between the required strength (168 kips) and the design strength (214 kips), wecould try a smaller section, W12x35. If we did, it will turn out that this section is not satisfactory.Hence, we select W12x40 .



Example 16-5: Design of Braced Column, (?)

Design an economical W section to hold a factored load Pu of 795 kips. The column is 20 ft. longand is assumed to have pin-pin end conditions. The column is braced at midheight in the weak directionand is made of A36 steel.Solution:

1. Since the column is braced at midheight in the weak direction, the effective length is 10 ft. in theweak direction, and 20 ft in the strong one.2. Assuming a design stress φcFcr = 30 ksi and

Atrialg =PuφcFcr

=79530

= 26.5 in2

3. Select W14x90 as a trial member (A = 26.5 in2)

KLry

= (1.0)(10) ft(12) in/ft(3.7) in = 32.4

KLrx

= (1.0)(20) ft(12) in/ft(6.14) in = 39.1 ✛

4. Using Table 16.1, for a slenderness ratio of 39.1, the design stress is 28.23 ksi, thus the design capacityis

φcPn = φcFcrAg = (28.23) ksi(26.5) in2 = 748.1 k

5. Since this is slightly less than the required capacity (795 kips), we need a bigger section and tryW14x99

KLry

= (1.0)(10) ft(12) in/ft(3.71) in = 32.34

KLrx

= (1.0)(20) ft(12) in/ft(6.17) in = 38.9 ✛

6. Using Table 16.1, for a slenderness ratio of 38.9, the design stress is 28.27 ksi, thus the design capacityis

φcPn = φcFcrAg = (28.27) ksi(29.1) in2 = 822.6 k

7. Hence, select W14x99 .

Example 16-6: Design of a Column, (?)

Select the lightest W section of A36 steel to serve as a pinned-end main member column 16 ft. longto carry an axial compression load of 95 kips dead load and 100 kips live load in a braced structure, asshown below. Design and indicate the first three choices.

Hinged

Hinged

P

P

16’ 0"

A 36Steel



Section Area (sq.in.) KL/ry Fcr (kksi) φcPn (kips)W8 X 48 14.1 92 23.0 276 First choiceW10 X 49 14.4 76 26.6 326 Second choiceW12 X 50 14.7 98 21.7 271 Not goodW14 X 53 15.6 100 21.3 282 Third choice

Solution:

1. Compute factored load.

Pu = 1.2D+ 1.6L = 1.2(95) + 1.6(100) = 274 k

2. We simply summarize the calculations for this example

Example 16-7: Column Design Using AISC Charts, (?)

Select the lightest W section of A36 steel to serve as a main member 30 ft. long to carry compressionload of 50 kips dead load and 110 kips live load in a braced structure, as shown below. The member isassumed pinned at top and bottom and in addition has weak direction support at mid-height.

P P

P P

15’ 0"

15’ 0"

A 36Steel

K L /r governsy y y

shape ifPotential buckled

Bracing

K L /r governsshape ifPotential buckled

for y-axis bending onlyAssume hinged at mid-height

bendingbottom for x- and y-axisAssume hinged top and

x x x

(a) (b)

Solution:

1. Compute the factored loads:

Pu = 1.2D+ 1.6L = 1.2(50) + 1.6(110) = 236 k

2. Select from LRFD Column Load Tables. The effective length factors K for buckling in either thestrong or the weak direction equal unity; i.e., Kx = Ky = 1.0. Since the AISC Column Load tables arecomputed assuming (KL/r)y controls, enter these tables with the effective length (KL)y. Thus enterwith,

Pu = 236 k; (KL)y = 15 ft.

Starting with the shallow W8 sections and working towards the deeper sections, findW8 X 40 φcPn = 238 k rx/ry = 1.73W10 X 45 φcPn = 267 k rx/ry = 2.15W12 X 45 φcPn = 257 k rx/ry = 2.65



3. Since the actual support conditions are such that (KL)x = 2(KL)y, if rx/ry ≥ 2, weak axis controlsand tabular loads give the correct answer. Thus, W10 X 45 and W12 X 45 are both acceptable.4. Since rx/ry for W8 X 40 is less than 2, strong axis bending controls. The strength may be obtainedfrom the tables by entering with the equivalent (KL)y that corresponds to (KL)x = 30ft.5. For equal strength with respect to the x and y axes,

(KL)xrx

=(KL)yry

(16.9-a)

(KL)y =(KL)xrx/ry

=30

1.73= 17.3 ft (16.9-b)

For (KL)y = 17.3 ft., the W8 X 40 has a design strength φcPn of only 208 kips; therefore it is notacceptable.6. Check of section. It is always advisable to make a final check of the apparent acceptable solution:W10X 45,

(KL/r)y = 15(12)/(2.10) = 89.6 (16.10-a)φcFcr = 20.1 ksi (16.10-b)

φcPn = φcFcrAg = (20.1)(13.3) = 267 k > 236 k√

(16.10-c)

Therefore use W10 X 45


Draft

Chapter 17

STEEL CONNECTIONS

17.1 Bolted Connections

1 Bolted connections, Fig. 17.1 are increasingly used instead of rivets (too labor intensive) and moreoften than welds (may cause secondary cracks if not properly performed).

17.1.1 Types of Bolts

2 Most common high strength bolts are the A325 and A490.

3 A325 is made from heat-treated medium carbon steel, min. F by ≈ 81− 92 ksi, F bu = 120 ksi

4 A490 is a higher strength manufactured from an alloy of steel, min. F by ≈ 115− 130 ksi, and F bu = 150ksi

5 Most common diameters are 3/4”, 7/8” for building constructions; 7/8” and 1” for bridges, Table 17.1.

Bolt Diameter (in.) Nominal Area (in2)5/8 0.30683/4 0.44187/8 0.60131 0.7854

1 1/8 0.99401 1/4 1.2272

Table 17.1: Nominal Areas of Standard Bolts

17.1.2 Types of Bolted Connections

6 There are two types of bolted connections:

Bearing type which transmits the load by a combination of shear and bearing on the bolt, Fig. 17.2.

Slip-critical transmits load by friction, Fig. 17.3. In addition of providing adequate at ultimate load,it must not slip during service loads.

7 Possible failure modes (or “limit states”) which may control the strength of a bolted connection areshown in Fig. 17.4.

Draft17–2 STEEL CONNECTIONS

Butt Splice Lap Splice

Axial Shear

Bracket ConnectionHanger Connection

TensionEccentric Shear

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

Standard Beam Connection Unstiffened Seat Connection Stiffened Seat Connection Structural Tee Connection

Shear and Tension

Figure 17.1: Examples of Bolted Connections

��

��

P

��

��

PP

P

P

��

��

��

�� P

PP

PP

Figure 17.2: Stress Transfer by Shear and Bearing in a Bolted Connection


Draft17.1 Bolted Connections 17–3

µT

µT

µT

PP

T

T

T

T P=

High-strength bolt

P

P

Figure 17.3: Stress Transfer in a Friction Type Bolted Connection

��

��

��

��

Shear Failure

PP

��

��

��

��

��

Tensile Failure

��

��

��

��

��

��

��

* Checked Through Design of Plates

Shear Failure of Plate *

Tensile Failure of Plate *

Bearing Failure of Plate

Bearing Failure

Bending Failure

Figure 17.4: Modes of Failure of Bolted Connections



17.1.3 Nominal Strength of Individual Bolts

Tensile Strength The nominal strength Rn of one fastener in tension is

Rn = F buAn (17.1)

where F bu is the tensile strength of the bolt, and An the area through the threaded portion, alsoknown as “tensile stress area”. The ratio of the tensile stress area to the gross area Ag ranges from0.75 to 0.79.

Shear Strength is given byRn = mAbFn (17.2)

where m is the number of shear planes, Fig. 17.5, Ab the gross area of the bolt, and Fn is thenominal shear strength of bolts

��

��

��

m=2

��

��

��

��

PP

m=1

Figure 17.5: Number of Shearing Planes m in Bolted Connections

Bearing Strength Rn is the force applied against the side of the hole to split or tear the plate, Fig.17.6. The larger the end distance L, the less likely that of having a splitting failure. Tearing willoccur along lines 1-1 and 2-2. In the most conservative case, α is taken as zero, thus

d/2

L

1

thickness t

P1

1

2

d

2 α

thickness t

Pα

Resisting Shear Stresses

Failure Line

Figure 17.6: Bearing Strength Related to End Distance

Rn = 2t[L− d

2

]τpu (17.3)

where:τpu is the shear strength of the plate ≈ 0.7FuFu is the tensile strength of the plate materiald is the nominal bolt diameter.

Thus this equation transforms into

Rn = 2t[L− d

2

](0.70Fu) (17.4-a)

= 1.40Fudt[L

d− 1

2

](17.4-b)

which can be approximated as

Rn = Fudt[L

d

]= LtFu (17.5)

which applies to the bolt closest to the edge for the design of single bolt connection, or two ormore in line force, when the end distance is less than 1.5d.



17.1.4 Concentric Loads (Tension Connections)

17.1.4.1 AISC Requirements

8 AISC RFD requirements for fasteners must satisfy

φRn ≥ ΣαiQi (17.6)

where φ is the strength resistance factor, Rn the nominal resistance (strength), αi the load magnificationfactor, and Qi the actual loads.

9 The most important relevant requirements are summarized below:

Shear Strength

No threads in shear plane (LRFD-J3.3)

φRn = φ(0.6F bu)mAb (17.7)

where φ = 0.65, and m the number of shear planes, A− b the gross-sectional area across theunthreaded shank of the bolt.

Threads in shear plane (LRFD-J3.3)

φRn = φ(0.45F bu)mAb (17.8)

where φ = 0.65, and m the number of shear planes, Ab the gross-sectional area across theunthreaded shank of the bolt.

Tension Strength (LRFD-J3.3)

φRn = φF bu(0.75)Ab (17.9)

where φ = 0.75.

Bearing Strength (LRFD-J3.6). Note that for Bearing strength, Fu is the one of the plate and notthe bolt.

1. Usual conditions (standard holes or short-slotted holes, end distance not less than 1.5d, boltspacing center to center not less than 3d, and with two or more bolts in the line of force:

φRn = φ(2.4dtFu) (17.10)

where φ=0.75, d is the nominal diameter of the bolt (not at threads), t thickness of connectedpart.

2. Long-slotted holes perpendicular to the direction of load transmission, end distance not lessthan 1.5d, bolt center-to-center spacing not less than 3d and with two or more bolts in lineof force (LRFD J3-1b)

φRn = φ(2.0dtFu) (17.11)

where φ=0.75, d is the nominal diameter of the bolt (not at threads), t thickness of connectedpart.



3. For the bolt closest to the edge and when Eqs 17.10 and 17.11 are not satisfied, then (LRFDJ3-1c)

φRn = φLtFu (17.12)

where L is the end distance of force, from the center of a standard hole or oversized hole, orfrom the mid-width of a slotted hole, to an edge of a connected part.

Minimum Spacing of bolts in the direction of transmitted forces must be at least 3 diameters whenEq. 17.10 and 17.11 are used. Under all other conditions Eq. 17.5 applies. Solving for L from thecenter of the fastener to the edge of the adjacent hole

L ≥ RnFut

(17.13)

then adding the radius dh/2, and replacing Rn by P/φ, gives the minimum center to center spacing

Spacing ≥ P

φFut+dh2

(17.14)

where φ = 0.75

Minimum End Distance When Eq. 17.10 and 17.11 are used, then the minimum end distances mustbe at least 1 1/2 diameters. If higher bearing strengths are used, then Eq. 17.5 gives

L ≥ RnFut

(17.15)

thus LRFDJ3.10 results in

L ≥ P

φFut(17.16)

where φ=0.75

In all cases F bu is the tensile strength of the bolt (120 ksi for A325; 150 ksi for A490).

17.1.4.2 Examples

Example 17-1: Analysis of Bolted Connections (Salmon and Johnson 1990)

Compute the tensile service load capacity for the bearing-type connection shown below if a) the boltthreads are excluded from the shear plane and b) the bolt threads are included in the shear plane. 7/8in diameter A325 bolts in standard holes, and A572 Grade 50 steel plates are used. LL=3 DL

��

��

��

��

��

��

��

�� T

T

T T

Plates 5/8 x 6

1 1/2"

3"1 1/2"

7/8" hole A325 Bolts

A572 Grade 50 Steel Plates



Solution:

1. Determine the strength based on tension members:

Ag = 6(0.625) = 3.75 in2 (17.17-a)

An =[6− 2

(78

+18

)](0.625) = 2.50 in2 (17.17-b)

Ae = UAn = 1(2.50) = 2.50 in2 (17.17-c)φTn = φFyAg = (0.9)(50)(3.75) = 169 k (17.17-d)φTn = φ FuAe = (0.75)(65)(2.5) = 122 k ✛ (17.17-e)

2. Shear strength

φRn = φ(0.60F bu)mAb (17.18-a)= 0.65(0.60)(120)(1)(0.6013) = 28.1 k/bolt (17.18-b)

3. Bearing strength for 1.5d end distance and 3d center to center spacing is

φRn = φ(2.4Fudt) (17.19-a)= 0.75(2.4)(65)(0.875)(0.625) = 64 k/bolt (17.19-b)

note that for the bearing strength, Fu is the ultimate strength of the plate.

4. Since bearing strength is larger than shear strength, shear controls and the total design strengthbased on the fasteners is

φTn = No. of fasteners Rn = 4(28.1) = 112 k (17.20)

5. The single shear design strength (112 kip) is smaller than the plate design strength (122 kip) thenφTn=112 kips

6. The end distance L from the center of the hole to the end of the plate should be, Eq. 17.16

L ≥ P

φFut=

28.10.75(65)(0.625)

= 0.92 in√

(17.21)

7. Shear block failure was not checked, it should have been

8. The design strength is related to the service load

φTn ≥ Tu (17.22-a)Tu = 1.2D + 1.6L (17.22-b)

112 = 1.2D + 1.6(3D) = 6.0D (17.22-c)D = 18.7 k (17.22-d)L = 3D = 56.1 k (17.22-e)

Tmax = D + L = 18.7 + 56.1 = 74.8 k (17.22-f)

9. If threads are included in the shear plane, the design strength per bolt in single shear is

φRn = φ(0.45F bu)mAb (17.23-a)= 0.65(0.45)(120)(1)(0.6013) = 21.1 k/bolt (17.23-b)



10. Strength in single shear governs also

φTn = No. of fasteners Rn = 4(21.1) = 84.4 k (17.24-a)84.4 = 1.2D+ 1.6(3D) = 6.0D (17.24-b)D = 14.1 k (17.24-c)L = 3D = 42.3 k (17.24-d)

Tmax = D + L = 14.1 + 42.4 = 56.4 k (17.24-e)

Example 17-2: Design of Bolted Connections (Salmon and Johnson 1990)

Determine the number of 3/4 in diameter A325 bolts in standard size holes required to carry 7 kipsdead load and 43 kips live load on the plates shown below using A36 steel. Assume the portion of thedouble lap splice is a bearing-type connection with threads excluded from the shear plane and a doublerow of bolts.

T

TT

T/2

T/2

1/4X10 Plates

10"

Solution:

1. The center plate controls the tension strength

Ag = (10)(0.25) = 2.50 in2 (17.25-a)

An =[10− 2

(34

+18

)]0.25 = 2.06 in2 (17.25-b)

Ae = UAn = (1)(2.06) = 2.06 in2 (17.25-c)φTn = φFyAg = 0.9(36)(2.50) = 81 k ✛ (17.25-d)φTn = φFuAe = 0.75(58)(2.06) = 90 k (17.25-e)Tu = 1.2D+ 1.6L = 1.2(7) + 1.6(43) = 77 k < φTn

√(17.25-f)

2. The shear strength of each A325 bolt is

φRn = φ(0.60F bu)mAb = 0.65(0.60)(120)(2)(0.4418) = 41.4 k/bolt (17.26)

3. The design strength in bearing on the 1/4 in plate is

φRn = φ(2.4Fudt) = 0.75(2.4)(58)(0.75)(0.25) = 19.6 k/bolt ✛ (17.27)

4. Thus the total number of bolts required is

No. of bolts =77

19.6= 3.9 (17.28)


Draft17.2 Welded Connections 17–9

5. The end distance is obtained from Eq. 17.16

P =774

= 19.3 k (17.29-a)

L ≥ P

φFut=

19.30.75(58)(0.25)

= 1.77 in (17.29-b)

6. Select the bolt arrangement shown below

��

��

��

��

T

2" 3"

2"

6"

2"

a

bc

ef

d

7. We must check shear rupture since the plate is a thin one and heavily loaded

Shear Yielding-Tension Fracture Tn = 0.60FyAvg + FuAnt (17.30-a)= 0.60(36)(10)0.25 + 58(4− 13/16)0.25 (17.30-b)= 54 + 46 = 100 k (17.30-c)

Shear Fracture-Tension Yielding Tn = 0.60FuAns + FyAtg (17.30-d)= 0.60(58)[10− 3(13/16)]0.25 + 36(4)0.25(17.30-e)= 66 + 36 = 102 k ✛ (17.30-f)

(17.30-g)

Note that we have used 3/4 + 1/16 for the hole diameter instead of 3/4 + 1/8

8. The design strength φTn givesφTn = 0.75(102) = 77 k (17.31)

which equals the factored load Tu = 77 kips that must be supported.

9. Use 2 in. edge distance along the sides, 3 in. longitudinal spacing, and 2 in. end distance, with 4A325 bolts.

17.1.5 Eccentric Connection (Shear Connections)

17.2 Welded Connections

10 Basic types of welds are shown in Fig. 17.7

t

L

o45 o

45t

b

t

L

Figure 17.7: Basic Types of Weld


Draft

Chapter 18

UNBRACED ROLLED STEELBEAMS

18.1 Introduction

1 In a previous chapter we have examined the behavior of laterally supported beams. Under thoseconditions, the potential modes of failures were either the formation of a plastic hinge (if the section iscompact), or local buckling of the flange or the web (partially compact section).

2 Rarely are the compression flange of beams entirely free of all restraint, and in general there are twotypes of lateral supports:

1. Continuous lateral support by embedment of the compression flange in a concrete slab.

2. Lateral support at intervals through cross beams, cross frames, ties, or struts.

3 Now that the beam is not laterally supported, we ought to consider a third potential mode of failure,lateral torsional buckling.

18.2 Background

4 Whereas it is beyond the scope of this course to derive the governing differential equation for flexuraltorsional buckling (which is covered in either Mechanics of Materials II or in Steel Structures), we shallreview some related topics in order to understand the AISC equations later on

5 There are two types of torsion:

Saint-Venant’s torsion: or pure torsion (torsion is constant throughout the length) where it is as-sumed that the cross-sectional plane prior to the application of torsion remains plane, and onlyrotation occurs.

Warping torsion: out of plane effects arise when the flanges are laterally displaced during twisting.Compression flange will bend in one direction laterally while its tension flange will bend in another.In this case part of the torque is resisted by bending and the rest by Saint-Venant’s torsion.

6 With reference to Fig. 18.1, the governing torsional differential equation is

Insert Fig. 9.4.1 from Steel Structures by Salmon and Johnson

Figure 18.1: I Shaped Beam in Slightly Buckled Position

ECwd4φ

dz4−Gj d

2φ

dz2− M

20

EIyφ = 0 (18.1)

Draft18–2 UNBRACED ROLLED STEEL BEAMS

where:Cw Torsional warping constantE Elastic modulusG Shear modulusJ Saint-Venant Torsional constantM0 Applied moment in the yz planeCw and J are section properties tabulated in the AISC manual.

7 Solution of this equation is

Mcr = Cbπ

L

√(πE

L

)2CwIy + EIyGJ (18.2)

where Cb is a factor to account for moment gradient.

18.3 AISC Equations

18.3.1 Dividing values

8 We identify three failure modes depending on Lb which is the distance between lateral supports (orunbraced length)

Conditions Failure ModeLb < Lp “very short” Plastic hingeLp < Lb < Lr “short” Inelastic lateral torsional bucklingLr < Lb “long” Elastic lateral torsional buckling

and

Lp =300√Fy, ksi

ry (18.3)

Lr =ryX1

Fyf − Fr

√1 +

√1 +X2 (Fyf − Fr)2 (18.4)

where:Lp Limit dividing line for plastic hinge formationLr Dividing line between inelastic and elastic lateral torsional bucklingry Radius of gyration with respect to the “weak” axis, [in.]Fyw Yield stress of the flange, [ksi]Fr Compressive residual stress, usually taken as 10 ksi. for rolled sections,

and 16.5 ksi for welded shapes.and

X1 =π

Sx

√EGJA

2(18.5-a)

X2 = 4CwIy

(SxGJ

)2(18.5-b)

are not really physical properties but a combination of cross-sectional ones which simplifies the writingof Eq. 18.4. Those values are tabulated in the AISC manual.

9 The flexural efficiency of the member increases when X1 decreases and/or X2 increases.

18.3.2 Governing Moments

1. Lb < Lp: “very short” Plastic hinge

Mn =Mp = ZxFy (18.6)


Draft18.3 AISC Equations 18–3

2. Lp < Lb < Lr: “short” inelastic lateral torsional buckling

Mn = Cb

[Mp − (Mp −Mr)

(Lb − LpLr − Lp

)]≤Mp (18.7)

and

Cb = 1.75 + 1.05(M1

M2

)+ 0.3

(M1

M2

)2≤ 2.3 (18.8)

where M1 is the smaller and M2 is the larger end moment in the unbraced segment M1M2

is negativewhen the moments cause single curvature (i.e. one of them is clockwise, the other counterclockwise),hence the most severe loading case with constant M gives Cb = 1.75− 1.05 + 0.3 = 1.0.

Mr is the moment strength available for service load when extreme fiber reach the yield stress Fy;

Mr = (Fy − Fr)Sx (18.9)

3. Lr < Lb “long” elastic lateral torsional buckling, and the critical moment is the same as in Eq.18.2

Mcr = Cbπ

Lb

√(πE

Lb

)2CwIy + EIyGJ ≤ CbMr and Mp (18.10)

or if expressed in terms of X1 and X2

Mcr =CbSxX1

√2

Lbry

√√√√1 +X21X2

2(Lbry

)2 ≤ CbMr (18.11)

10 Fig. 18.2 summarizes the governing equations.

✲

✻Mn

Lb

Mp1

�

2�

❅❅

❅❅

❅❅ 3

�

✲✛ ✲✛Lateral Torsional Buckling

Mr

Lr =ryX1

Fyf−Fr

√1 +

√1 +X2

(Fyf − Fr

)2Lp =300√

Fyry

PlasticAnalysis

Mp = ZxFy

InelasticMn = Cb

[Mp − (Mp −Mr)

(Lb−LpLr−Lp

)]

ElasticMcr = CbSxX1

√2

Lbry

√1 +

X21X2

2(Lbry

)2

Figure 18.2: Nominal Strength Mn of “Compact” Sections as Affected by Lateral-Torsional Buckling



18.4 Examples

18.4.1 Verification

Example 18-1: Adequacy of an unbraced beam, (?)

Verify the adequacy of a 15 ft W12X87 beam, subjected to a uniform load of 3k/ft and which isbraced only at the ends. The load is 60% live and 40% dead and unfactored. Steel is A36 and Cb = 1.0.Solution:

1. The factored load and moments are

wu = 1.2(0.4)(3) + 1.6(0.6)(3) = 4.32 k/ft (18.12-a)

Mu =wuL

2

8=

(4.32)(15)2

8= 121.5 k.ft (18.12-b)

2. Next we check if the W12X87 is compact, From Eq. 14.12

bf2tf

= 7.5 <65√36︸︷︷︸λp

= 10.8√

(18.13-a)

hctw

= 18.9 <640√36︸︷︷︸λp

= 106.√

(18.13-b)

Note that those values are taken from the AISC manual.3. The unbraced length is Lb=15 ft.4. From AISC manual,

ry = 3.07 in (18.14-a)X1 = 3, 880 ksi (18.14-b)X2 = 0.000586 (18.14-c)

5. Determine Lp from Eq. 18.3

Lp =300√Fyry (18.15-a)

=300√36

(3.07) = 153.5 in = 12.8 ft (18.15-b)

6. Determine Lr from Eq. 18.4

Lr =ryX1

Fyf − Fr

√1 +

√1 +X2 (Fyf − Fr)2 (18.16-a)

=(3.07) in(3, 880) ksi

(36− 10) ksi

√1 +

√1 + 0.000586 (36− 10)2 (18.16-b)

= 676.6 in = 56.4 ft (18.16-c)



7. Hence Lp < Lb < Lr, and failure is caused by inelastic lateral torsional buckling, and we must useEq. 18.7

Mn = Cb

[Mp − (Mp −Mr)

(Lb − LpLr − Lp

)]≤Mp (18.17-a)

Mp = ZxFy = (36) ksi(132) in31

(12) in/ft= 396 k.ft (18.17-b)

Mr = (Fy − Fr)Sx = (36− 10)(118)1

(12)= 255.7 k.ft (18.17-c)

Mn = 1.0[396− (396− 255.7)

(15− 12.8

56.4− 12.8

)]= 388.9 k.ft ≤ 396 k.ft

√(18.17-d)

8. The moment capacity will then be

φbMn = (0.9)(388.9) k.ft = 350 k.ft > 121.5 k.ft√

(18.18)

Example 18-2: Adequacy of an unbraced beam, II (?)

Check the design of a 25 ft. long W10X33 beam which is subjected to a factored load (including selfweight) of 1.2 kips/ft. The beam is unbraced and made from A242 steel (Fy=50 ksi). Assume Cb = 1.0.Solution:

1. The factored moment is

Mu =wuL

2

8=

(1.2)(25)2

8= 93.75 k.ft (18.19)

2. Check if the W10X33 is compact, From Eq. 14.12

bf2tf

= 9.1 <65√50︸︷︷︸λp

= 9.19√

(18.20-a)

hctw

= 27.1 <640√50︸︷︷︸λp

= 90.5√

(18.20-b)

3. The unbraced length is Lb=25 ft.4. From AISC manual,

ry = 1.94 in (18.21-a)X1 = 2, 710 ksi (18.21-b)X2 = 0.00251 (18.21-c)

5. Determine Lp from Eq. 18.3

Lp =300√Fyry (18.22-a)

=300√50

(1.94) = 82.3 in = 6.85 ft (18.22-b)



6. Determine Lr from Eq. 18.4

Lr =ryX1

Fyf − Fr

√1 +

√1 +X2 (Fyf − Fr)2 (18.23-a)

=(1.94)(2, 710)

(50− 10)

√1 +

√1 + 0.00251 (50− 10)2 (18.23-b)

= 236.5 in = 19.7 ft (18.23-c)

7. Since Lb > Lr, failure is caused by elastic lateral torsional buckling, and we must use Eq. 18.8

Mcr =CbSxX1

√2

Lbry

√√√√1 +X21X2

2(Lbry

)2 (18.24-a)

=(1.0)(35.0)(2, 710)

√2

(25)(12)1.94

√√√√1 +(2, 710)2(0.00251)

2((25)(12)1.94

)2 (18.24-b)

= 1, 021 k.in = 85.1 k.ft (18.24-c)

8. The moment capacity will then be

φbMn = (0.9)(85.1) k.ft = 76.6 k.ft < 93.75 k.ft NG (18.25)

Hence the section is not adequate.

18.4.2 Design

11 Seeking a design where failure is through a plastic hinge, the objective is to have Lb < Lp. This canbe achieved by setting Lb = Lp, or from Eq. 18.3

rymin =Lb√Fy

300(18.26)

and from Eq. 18.6

Zxreq =Mn

Fy(18.27)

12 If Lb is so large that the previous design objectives can not be met, then the bam will fail byeither inelastic or elastic lateral torsional buckling. Recalling that the limiting cases for the former isLp < Lb < Lr, and that the corresponding limiting moments Mp and Mr are given by Eq. 18.6 and 18.9respectively, then the upper and lower limits would be

Zxreq <Mn

Fy(18.28-a)

Sxreq >Mn

Fy − Fr (18.28-b)

Hence, if Lb is close to Lp, then we should select a section using Zxreq , and if Lb is close to Lr, thenwe select a section close to Sxreq .

13 Alternatively, one can use the AISC design charts. In those charts, the design moment φbMn isplotted in terms of Lb for a given Fy and Cb = 1.0. In those charts, the most economical section by



weight are differentiated from the others (solid versus dashed lines), and limiting Lp and Lr values arehighlighted by closed and open circles.

14 To use those charts, one would enter Lb and φbMn, where those two values intersect, then any beamlisted to the right or above will satisfy the necessary requirements (assuming that Cb = 1.0).

15 Note that since all of the strength equations are linear in Cb, we can use Lb/Cb as a substitute of Lbwith Cb = 1

Example 18-3: Design of Laterally Unsupported Beam, (?)

Select an economical W section for the beam shown in Fig. 18.3. Lateral support is provided at thevertical supports, concentrated load points, and at the end of the cantilever. The 26-kip load contains6 kips dead load and the 11.5-kip load includes 4 kips dead load; the remainder is live load. Use A36steel. Use W24 section.

460

70

353

101

24’ 28’ 21’A B C

DL: 6LL: 20

DL 4LL: 7.5

LS LS LS LS

LL on 52 ft span, not on cantilever

LL on cantilever, not on 52 f span

Factored bending moment envelope due to superimposed load

+

-

Figure 18.3: Design of Laterally Unsupported Steel Beam

Solution:

1. The moment envelope resulting from the factored load is first obtained. It contains two loadingscenario; a) DL+LL on the 52-ft span, and no LL on the cantilever; and b) DL+LL on the cantileverand no LL on the 52 ft span:

Wu1 = 1.2(6) + 1.6(20) = 39.2 k (18.29-a)Wu2 = 1.2(4) + 1.6(7.5) = 16.8 k (18.29-b)

2. For the first loading case, the maximum factored moment at the cantilever support and under thepoint load

Mu = (1.2)(4)(21) = 101 k.ft (18.30-a)

Mu = (39.2)(24)(58)12− (101)

2452

= 460 k.ft (18.30-b)



3. For the second load case, the maximum factored moment at the cantilever support is

Mu = (16.8)(21) = 353 k.ft (18.31)

4. Three laterally unbraced segments must be considered since each length is different and is subjectedto a different maximum factored moment Mu. By inspection, segment A controls over segment C; boththe moment and the unbraced length are larger on segment A, and the moment gradient is the same(i.e. Cb=1.75 for both).5. Segment B has a larger Cb

Cb = 1.75 + 1.05(M1

M2

)+ 0.3

(M1

M2

)2≤ 2.3 (18.32-a)

CAb = 1.75 + 1.05(

0460

)+ 0.3

(0

460

)2= 1.75 (18.32-b)

CBb = 1.75 + 1.05(

+101+460

)+ 0.3

(+101+460

)2= 2.0 ≤ 2.3 (18.32-c)

which is beneficial, but also a larger Lb, hence we should design both segment A and B.6. For segment A, Mu=460 ft-kips, and Lb= 24 ft. The required Mn is

Mn =Mu

φb=

4600.9

= 511 k.ft (18.33)

7. Aiming for a plastic failure, we use Eq. 18.26 to determine rymin and Zxmin

rymin =Lb√Fy

300=

(24)(12)√

36300

= 5.76 in (18.34-a)

Zxmin =Mn

Fy=

(511) k.ft(12) in/ft

(36) ksi= 170 in3 (18.34-b)

8. From the AISC manual we have the following

Section ry Zx Sx Lp Lr RemarksW24x492 3.41 1,550 1,290 Largest W24 section,

ry NG and section too stiff (very large Zx)W24x76 1.92 200. 176. 8.0 23.4 Zx OK but ry NGW24x68 1.87 177. 154. 7.8 22.4 Zx OK but ry NGW24x55 1.34 134 114. 5.6 16.5 Smallest W24 section

Hence it is quite clear that failure will have to be through inelastic or elastic torsional buckling.9. We try W24x76.10. Is the section compact?

bf2tf

= 6.66 <65√36︸︷︷︸λp

= 10.8√

(18.35-a)

hctw

= 49 <640√36︸︷︷︸λp

= 106√

(18.35-b)

Then we can proceed,11. From Eq. 18.3 and 18.4:

Lp =300√Fyry =

112

300√36

1.92 = 8.0 ft (18.36-a)

Lr =ryX1

Fyf − Fr

√1 +

√1 +X2 (Fyf − Fr)2 (18.36-b)



=112

(1.92)(1, 760)36− 10

√1 +

√1 + (0.0186) (36− 10)2 (18.36-c)

= 23.4 ft (18.36-d)

12. Since Lr < Lb, failure is governed by elastic lateral torsional buckling, and we use Eq. 18.11

Mr = (Fy − Fr)Sx = (36− 10)(176)1

(12)= 381. k.ft (18.37-a)

CbMr = (1.75)(381) = 667 k.ft (18.37-b)

Mp = ZxFy =(200)(36)

12= 600 k.ft (18.37-c)

Lbry

= =(24)(12)

1.92= 150 (18.37-d)

Mcr =CbSxX1

√2

Lbry

√√√√1 +X21X2

2(Lbry

)2 ≤ CbMr (18.37-e)

=1

(12)(1.75)(176)(1, 760)

√2

150

√1 +

(1, 760)2(0.0186)2(150)2

(18.37-f)

= 643.14 k.ft ≤ CbMr but > Mp (18.37-g)⇒Mn = Mp = 600 k.ft (18.37-h)φbMn = (0.9)(600) = 540 k.ft

√(18.37-i)

13. Hence the W24x76 is a satisfactory design.14. In an attempt to automate the design procedure through a spread-sheet, and seek the most eco-nomical design (smallest weight), we consider table 18.1. and note that W24x68 is the most economical

Section Lb/ry Lp Lr Mp Mr CbMr Eq. 18.7 Mine Eq. 18.8 Mel. φbMn

ft. ft. ft-kips ft-kips ft-kips ft-kips ft-kips ft-kips ft-kips ft-kipsW24.x76. 150.00 8.0 23.4 600. 381. 667. 0. 0. 643. 600. 540.W24.x68. 154.01 7.8 22.4 531. 334. 584. 0. 0. 523. 523. 471.W24.x62. 208.70 5.8 17.2 459. 284. 497. 0. 0. 298. 298. 268.W24.x55. 214.93 5.6 16.6 402. 247. 432. 0. 0. 239. 239. 215.W21.x83. 157.38 7.6 24.9 588. 371. 648. 669. 588. 0. 0. 529.W21.x73. 159.12 7.5 23.5 516. 327. 573. 0. 0. 555. 516. 464.W21.x68. 160.00 7.5 22.8 480. 303. 531. 0. 0. 491. 480. 432.W21.x62. 162.71 7.4 21.7 432. 275. 482. 0. 0. 414. 414. 372.W18.x97. 108.68 11.0 38.1 633. 407. 713. 918. 633. 0. 0. 570.W18.x86. 109.51 11.0 35.5 558. 360. 629. 792. 558. 0. 0. 502.W18.x76. 110.34 10.9 33.3 489. 316. 554. 679. 489. 0. 0. 440.W16.x100. 114.74 10.5 42.1 594. 379. 664. 879. 594. 0. 0. 535.W16.x89. 115.66 10.4 38.6 525. 336. 588. 759. 525. 0. 0. 473.W16.x77. 116.60 10.3 34.9 450. 290. 508. 632. 450. 0. 0. 405.W14.x109. 77.21 15.5 62.7 576. 375. 656. 945. 576. 0. 0. 518.W14.x99. 77.63 15.5 58.2 519. 340. 595. 846. 519. 0. 0. 467.W14.x90. 77.84 15.4 54.1 471. 542. 762. 762. 471. 0. 0. 424.W12.x120. 92.01 13.0 75.5 558. 353. 618. 914. 558. 0. 0. 502.W12.x106. 92.60 13.0 67.2 492. 314. 550. 798. 492. 0. 0. 443.W12.x96. 93.20 12.9 61.4 441. 284. 497. 709. 441. 0. 0. 397.W10.x112. 107.46 11.2 86.4 441. 273. 478. 722. 441. 0. 0. 397.W10.x100. 108.68 11.0 77.4 390. 243. 425. 632. 390. 0. 0. 351.W8.x67. 135.85 8.8 64.0 211. 131. 229. 330. 211. 0. 0. 190.W8.x58. 137.14 8.8 56.0 179. 113. 197. 276. 179. 0. 0. 161.W6.x25. 189.47 6.3 31.3 57. 36. 63. 74. 57. 0. 0. 51.W6.x20. 192.00 6.3 25.6 45. 29. 51. 53. 45. 0. 0. 40.

Table 18.1: Unbraced Beam Design

design. Should we increase the depth, then the weight increases (as expected for a beam).



15. Next, we must consider segment B, it has a steeper moment gradient than segment A, i.e. Cb > 1.75

Cb = 1.75 + 1.05(

+101+460

)+ 0.3

(+101+460

)2= 2.0 ≤ 2.3 (18.38)

however the laterally unbraced length of 28 ft is now longer.16. Check the W24x68

Lr =ryX1

Fyf − Fr

√1 +

√1 +X2 (Fyf − Fr)2 (18.39-a)

=112

(1.87)(1, 590)36− 10

√1 +

√1 + (0.0290) (36− 10)2 (18.39-b)

= 22.4 ft (18.39-c)Mp = φbZxFy (18.39-d)

= (177)(36)112

= 531 k.ft (18.39-e)

Since Lr < Lb, the beam fails by linear elastic torsional buckling and

Mr = (Fy − Fr)Sx = (36− 10)(154)1

(12)= 333.7 k.ft (18.40-a)

CbMr = (2.0)(333.7) = 667 k.ft (18.40-b)

Mcr =CbSxX1

√2

Lbry

√√√√1 +X21X2

2(Lbry

)2 ≤ CbMr (18.40-c)

Lbry

= =(28)(12)

1.87= 154. (18.40-d)

=1

(12)(2.0)(154)(1, 590)

√2

154

√1 +

(1, 590)2(0.0290)2(154)2

(18.40-e)

= 468 k.ft ≤ CbMr and > Mp (18.40-f)

17. We observe that Mcr = 468 k.ft for segment B, and 523 k.ft for segment A. Hence, segment Bcontrols.18. Checking for the ultimate moment

φbMn = (0.9)(468) = 421 < Mu (460) (18.41-a)

19. Hence, we must increase the section to W24x76. From above Lr = 23.4 ft. since Lr < Lb, failure isgoverned by elastic lateral torsional buckling, and we use Eq. 18.11

Mr = (Fy − Fr)Sx = (36− 10)(176)1

(12)= 381. k.ft (18.42-a)

CbMr = (2.0)(381) = 762 k.ft (18.42-b)

Mp = ZxFy =(200)(36)

12= 600 k.ft (18.42-c)

Lbry

= =(28)(12)

1.92= 175 (18.42-d)

Mcr =CbSxX1

√2

Lbry

√√√√1 +X21X2

2(Lbry

)2 ≤ CbMr (18.42-e)

=1

(12)(2.0)(176)(1, 760)

√2

175

√1 +

(1, 760)2(0.0186)2(175)2

(18.42-f)

= 581 k.ft ≤ CbMr and > Mp (18.42-g)φbMn = (0.9)(581) = 523 k.ft

√(18.42-h)


Draft18.5 Summary of AISC Governing Equations 18–11

20. Note that Mcr is 581 for segment B, and was 600 for segment A.21. Use W24x76

18.5 Summary of AISC Governing Equations

Table 18.2 summarizes the AISC design equations covered so far.

Braced Beamsbf2tf≤ λp

Compact Mn = ZFyhctw≤ λp

λp <bf2tf≤ λr

Partially compact Mn =Mp − (Mp −Mr)(λ−λpλr−λp ) ≤Mp

λp <hctw≤ λr

Unbraced BeamsPlastic Hinge Lb < Lp Mn =Mp = ZxFyInelastic lateral torsional buckling Lp < Lb < Lr Mn = Cb

[Mp − (Mp −Mr)

(Lb−LpLr−Lp

)]≤Mp

Elastic lateral torsional buckling Lr < Lb Mcr = CbSxX1√2

Lbry

√1 + X2

1X2

2(Lbry

)2 ≤ CbMr

ColumnsElastic Buckling λc ≤ 1.5 Fcr = 0.658λ

2cFy

Inelastic Buckling λc > 1.5 Fcr = 0.877λ2cFy

Table 18.2: Summary of AISC Governing Equations


Draft

Chapter 19

Beam Columns, (Unedited)

UNEDITEDExamples taken from Structural Steel Design; LRFD, by T. Burns, Delmar Publishers, 1995

19.1 Potential Modes of Failures

19.2 AISC Specifications

PuφcPn

+89

(Mux

φbMnx+

Muy

φbMny

)≤ 1.0 if

PuφcPn

≥ .20 (19.1)

Pu2φcPn

+Mux

φbMnx≤ 1.0 if

PuφcPn

≤ .20 (19.2)

19.3 Examples

19.3.1 Verification

Example 19-1: Verification, (?)

A W 12 × 120 is used to support the loads and moments as shown below, and is not subjectedto sidesway. Determine if the member is adequate and if the factored bending moment occurs aboutthe weak axis. The column is assumed to be perfectly pinned (K = 1.0) in both the strong and weakdirections and no bracing is supplied. Steel is A36 and assumed Cb = 1.0.

Draft19–2 Beam Columns, (Unedited)

Solution:

1. Calculate the slenderness ratios in both the weak and strong directions (even though when unbracedin both directions, the weak axis will of course be critical).

Klxrx

= (1.0)(15) ft(12) in/ft(5.51) in = 32.67

Klyry

= (1.0)(15) ft.(12) in/ft)(3.13) in = 57.50 ✛

2. Using Table 16.1, we can find the design axial stress (φcFcr) as 25.71 ksi. And, multiplying thatwith the area of the section, we can find the design axial capacity (φcPn) as follows:

φcPn = φcFcrAg= (25.71) ksi(35.3) in2 = 907.6 k

3. Since the factored axial load is 400 kips, the ratio of PuφcPn

is as follows:

(400) k

(907.6) k= .44 > .20

√

therefore we must use Eq. as the beam-column analysis.

PuφcPn

+89

(Mux

φbMnx+

Muy

φbMny

)≤ 1.0

4. To start, calculate the design moment capacity (φbMn) of the W 12 × 120. The beam is compactand the unbraced length, lb, is 15 feet. This value of unbraced length is then compared to limitsreferred to as Lp and Lr to determine whether plastic behavior can be developed or whetherbuckling behavior controls the bending behavior. The values for Lp and Lr can be calculatedusing Eq. 6-1 and 6-2 respectively or they can be found in the beam section of the LRFD manual.In either case these values for a W 12 × 120 are:

Lp = 13 ft

Lr = 75.5 ft



5. Since our unbraced length falls between these two values, the beam will be controlled by inelasticbuckling, and the nominal moment capacity Mn can be calculated from Eq. 6-5. Using thisequation we must first calculate the plastic and elastic moment capacity values, Mp and Mr.

Mp = FyZx = (36) ksi(186) in3 ft(12) in = 558 k.ft

Mr = (Fyw − 10) ksiSx(36− 10) ksi(163) in3 = 353.2 k.ft

6. Using Eq. 6-5 (assuming Cb is equal to 1.0) we find:

Mn = Cb[Mp − (Mp −Mr)]lb−LpLr−Lp

Mn = 1.0[558− (558− 453.2)] k.ft (15−13) ft(75.5−15) ft

= 551.4 k.ft

7. Therefore the design moment capacity is as follows:

φbMn = 0.90(551.4) k.ft = 496.3 k.ft

8. Now consider the effects of moment magnification on this section. Based on the alternative methodand since the member is not subjected to sidesway (Mlt = 0)

Mu = B1Mnt

B1 = Cm1−Pu

Pe

Cm = .60− .4M1M2

Cm = .60− .4−5050 = 1.0

Pu = 400 k

Pe = π2EAgKlr2 (Euler’s Buckling Load Equation)

Pe = π2(29,000 ksi)(35.3 in2)32.672 = 9, 456 k

9. Therefore, calculating the B1 magnifier we find:

B1 =1

1− (400) k(9,456) k

= 1.044

Calculating the amplified moment as follows:

Mu = B1Mnt

Mu = 1.044(50) k.ft = 52.2 k.ft

Therefore the adequacy of the section is calculated from Eq. as follows:

PuφcPn

+ 89

MuxφbMnx

≤ 1.094000 k(907.6) k + 8

9(52.2) k.ft(496.3) k.ft = .53 < 1.0

√

1 When dealing with unbraced frames, the procedure gets more involved since the calculation of sideswayeffects (sidesway moments and sidesway axial loads) is required to utilize the alternate method that weperform to estimate the secondary effects. This generally means that a single load case on an unbracedframe that causes sidesway must be analyzed as if it were two separate load cases—one that wouldhave a negligible sidesway effect (typically the vertical gravity loads) and one that would cause sidesway(typically the lateral loads). This is illustrated in Figure 19.1

2 As the forces and moments for each of these “broken-down” load cases are calculated for a member,the same basic procedure is then followed as it was in the first example.



Figure 19.1: Rigid Frame Subjected to Lateral Loading

3 The analysis of frames and frames subject to sidesway is not within the scope of this course.

4 You should be aware that as unbraced members are encountered, the B2 magnifier will become activeas the frame exhibits sidesway behavior. The following example illustrates this situation.

Example 19-2: 8.2, (?)

In the unbraced frame shown below, determine the adequacy of either W 14 × 90 columns under thefactored loads given. The steel is A36 and assume that the columns have a Ky = 1.0 and Kx = 1.5 Thelateral load is causing bending to occur about the x-axis.

Solution:

1. Typically, we would start this problem by separating the loads into those that cause sideswayeffects (the lateral load) and those that have little influence on such effects (the vertical loads).As shown below, the effects on the columns from the lateral load and the vertical load can becomputed separately by various standard techniques of analysis such as moment distribution.



2. From these techniques, we find that each column is under approximately 20 kips of factored axialload and a moment of 40 kip-feet in no sidesway case. And, they are under 20 kips of factoredaxial load (different directions) and a moment of 200 kip-feet (assuming pinned supports) in thesidesway case.

3. To begin the evaluation, calculate the column section’s design axial capacity φcPn. Calculatingthe slenderness ratios about both axis (although the weak axis will control in this case):

Klry

= 1.0(240) in(3.70) in = 64.8

Klrx

= 1.5(240) in(6.14) in = 58.6

4. Since the weak axis controls, find the design axial stress φcFcr that can be taken from Table 16.1.In this case the design axial stress corresponding to the critical slenderness ratio is 24.54 ksi. Thiswould make the design axial capacity of this section as follows:

φcPn = φcFcrAgφcPn = (24.54) ksi(26.5) in2) = 650.3 k

Pu = (20) k (no sidesway) + (20) k (sidesway) = 40 k

5. Since PuφcPn

= 40 k650.3 k = .061 < .20 Eq. controls and should be used to check the adequacy of the

section.

6. Calculate the design bending capacity of these sections about the strong axis. The unbraced lengthof this column is assumed to be 20 feet and the nominal moment capacity depends on what bendingbehavior controls in this case. Compare the unbraced length (20 ft.) to the Lp and Lr limits thatwere discussed in Chapter 6.

7. These values can be calculated from Equations 6-1 and 6-2 or taken from the LRFD manual. Sincethe Lp value for a W 14 × 90 is 15.4 feet and the Lr, value is 54.1 feet, the section’s capacity iscontrolled by Case 2 (compact beams exhibiting inelastic lateral torsional buckling).

8. The capacity of this section is found by calculating Eq. 6-5 (see Chapter 6) as follows and noticethat Cb is equal to 1.67 using Mmax = 40,Ma = 10,Mb = 20, and Mc = 30.

9. Before calculating the nominal moment capacity from Eq. 6-5, first calculate the plastic and elasticmoment capacity Mp and Mr.

Mp = FyZx = (36) ksi(157) in3 ft(12) in = 471 k.ft

Mr = (Fyw − 10 ksi)Sx= (36− 10) ksi(143) in3 = 309.8 k.ft



Calculating Eq. 6-5:

Mn = 1.67[(471) k.ft− [(471) k.ft(309.8) k.ft)((20− 15.4) ft(54.1− 15.4) ft)]= 791.5 k.ft

10. However because the nominal moment capacity cannot be greater than the plastic moment capacity,use 471 ft.-kips which is the plastic moment capacity. The design moment capacity is:

φbMn = 0.90(471) k.ft = 423.9 k.ft

11. After this we can calculate the magnification factors B1 and B2, remembering that B1 is used forthe nonsway effects and B2 for the sway effects. In this case, the axis of bending has been statedto be the x-axis and the effective length factor K was given as 1.5. Therefore for B1

B1 = Cm(1−Pu

Pe

Cm = .60− .4M1M2

Cm = .6 since M1M2

= 0

Pe = π2EAg

(Klr )2= π2E(26.5) in2

58.62

= 2206 k

Pu = 40 k

12. Therefore,

B1 =.6(

1− (40) k(2,206) k

) = .61 use 1.0

13. Calculating B2 from Eq. 8-5:

B2 = 1

1−ΣPuΣPe

ΣPu = 2(20) k = 40 k (for both columns in the story)ΣPe = (2, 206) k(2) columns = 4, 412 k

B2 = 11 − 40

4,412 = 1.01

14. Calculating the nominal moment capacity about the x-axis, Mus, from Equation 8-1 as follows:

Mu = B1Mnt +B2Mlt

Mux = 1.0(40) k.ft + 1.01(200) k.ft = 242 k.ft

15. Calculating the adequacy per Eq. :

Pu2φcPn

+ Mux

φbMnx= (40) k

(2)(650.3) k + (242) k.ft(423.9) k.ft

= .60 < 1.0

16. Therefore the section is good. Notice that the calculation of Pe for the B1 and B2 magnifiers isalways taken about the axis of bending, in this case the strong axis.

19.3.2 Design

5 The design of beam-columns is truly a trial and error procedure and depends a great deal on thedesigner’s experience. If the designer can select a good trial section at the very beginning, the workin obtaining a safe and economical member is greatly simplified although, in any case, the process willultimately converge on a solution.

6 The most popular method of selecting a beam-column is the equivalent axial load method. This methodreplaces the factored axial load and moment that are applied to the column with a fictitious concentric



axial load. This fictitious concentric axial load will be larger than the factored axial load but will produceapproximately the same maximum effect.

7 The following equation is used to convert the factored bending moment into an estimated axial load,P ′. The actual factored axial load Pu, is added to P ′ resulting in a fictitious axial load called theequivalent factored axial load Puequiv . This idea is shown below:

Pu + P ′ = Puequiv

wherePu actual factored axial loadP ′ estimated axial load caused by moment

Puequiv Equivalent factored axial load

8 Although there are equations that would estimate the equivalent factored axial load based on theassumptions used by the interaction formulas separately (Eq. and ), it is much faster to use the followingcombined approximation developed from these formulas. This approximation is as follows:

Puequiv = Pu +Muxm+MuymU (19.3)

In this formula, Pu is again the actual factored load expressed in kips andMux andMuy are the bendingmoments expressed in kip-feet.

9 If bending should occur about one axis only, the corresponding term regarding the other axis will thendrop out. The value of m is taken from Table 19.1.

Preliminary Beam-Column DesignFy = 36 ksi, Fy = 50 ksi

Values of mFy 36 ksi 50 ksi

KL(ft) 10 12 14 16 18 20 22 and 10 12 14 16 18 20 22 andover over

1st ApproximationAll 2.0 1.9 1.8 1.7 1.6 1.5 1.3 1.9 1.8 1.7 1.6 1.4 1.3 1.2

ShapesSubsequent Approximation

W4 3.1 2.3 1.7 1.4 1.1 1.0 0.8 2.4 1.8 1.4 1.1 1.0 0.9 0.8W6 3.2 2.7 2.1 1.7 1.4 1.2 1.0 2.8 2.2 1.7 1.4 1.1 1.0 0.9W8 2.8 2.5 2.1 1.8 1.5 1.3 1.1 2.5 2.2 1.8 1.5 1.3 1.2 1.1W8 2.5 2.3 2.2 2.0 1.8 1.6 1.4 2.4 2.2 2.0 1.7 1.5 1.3 1.2W10 2.1 2.0 1.9 1.8 1.7 1.6 1.4 2.0 1.9 1.8 1.7 1.5 1.4 1.3W12 1.7 1.7 1.6 1.5 1.5 1.4 1.3 1.7 1.6 1.5 1.5 1.4 1.3 1.2W14 1.5 1.5 1.4 1.4 1.3 1.3 1.2 1.5 1.4 1.4 1.3 1.3 1.2 1.2

Table 19.1: Values of m for use in Beam-Column Design Equation, (from a paper in AISC EngineeringJournal by Uang, Wattar, and Leet (1990)).

10 The value of U , which is the bending moment conversion factor, is taken from the column load tables thatare shown in Appendix C. (A full set of the column load tables can be found in the column section of the LRFDmanual as well).

11 The procedure using this approximate formula is an iterative process that begins by assuming the value of mfrom Table 8-1 and a typical value of U (usually about 2.0). After a section is selected, the section is checkedagainst the applicable AISC interactions formulas (Eq. and ). With successive trials the values of m and U canbe refined to a point where their values begin to stabilize.



Example 19-3: Design of Steel Beam-Column, (?)

Design the most economical W 12 section subjected to the loads and moments shown below. The value ofK = 1.0 in both strong and weak directions and steel is A36. Assume this member to be part of a frame that isnot subjected to sidesway.Solution:

Using the equivalent load formula and choosing the first trial values of m = 1.9 as suggested In Table 8-1 andU = 2.0 (approximate), the equivalent load is as follows:

Puequiv = Pu + Muxm + MuymUPuequiv = (300) k + (120) k.ft(1.9) + (80) k.ft(1.9)(2.0)

= 832 k

From Column Load tables found in Appendix C for Kl =12 ft. try a W 12 x 106 (with an axial design strengthof 853 kips)

Checking this section with the interaction formulas we find the following:

Since PuφPn

= (300) k(853) k

= .352 > .20, the Eq. will control.

Calculating the design moment capacity about each axis we find the following:

Lp = 13.0 ft

Since this value is greater than 12 ft. which is this W 12 x 106’s unbraced length, the section is able to developas plastic capacity. Therefore,

φbMnx = 0.90(36) ksi(164) in3 1

12 in/ft= 442.8 k.ft

φbMny = .90(36) ksi(75.1) in3 ft12 in = 202.8 k.ft

Calculating the magnifier, B1, similar to what was done previously, we would find the following:

Cmx = .6− .4−120120

= 1.0Cmy = 1.0( similarly)

Klrx

= 26.3(λc = .295)Klry

= 46.3(λc = .519)

Calculating the Euler buckling load as follows:

Pe =AgFy

λ2c

Pex = (31.2) in2(36) ksi.2952

Pex = (12, 907) k (based on λc = .295)

Pey = (31.2) in2(36) ksi.5192

Pey = (4, 162) k (based on λc = .519)B1x = 1

1− 300

12,907= 1.023

B1y = 11− 300

4,162= 1.078



This would make the magnified factored as shown below:

Mu = B1Mnt

Mux = (120) k.ft(1.023) = 122.76 k.ftMuy = (80) k.ft(1.078) = 86.24 k.ft

Applying AISC Eq. to the above information:((300) k

(853) k+

8

9

[(122.76) k.ft

(442.8) k.ft+

(86.24) k.ft

(202.8) k.ft

])= .976 < 1.0

√

which works well and therefore is a very good choice!


Draft

Part II

STRUCTURAL ENGINEERING II;CVEN 3235

Draft

Chapter 20

ARCHES and CURVEDSTRUCTURES

1 This chapter will concentrate on the analysis of arches.

2 The concepts used are identical to the ones previously seen, however the major (and only) differenceis that equations will be written in polar coordinates.

3 Like cables, arches can be used to reduce the bending moment in long span structures. Essentially,an arch can be considered as an inverted cable, and is transmits the load primarily through axialcompression, but can also resist flexure through its flexural rigidity.

4 A parabolic arch uniformly loaded will be loaded in compression only.

5 A semi-circular arch unifirmly loaded will have some flexural stresses in addition to the compressiveones.

20.1 Arches

6 In order to optimize dead-load efficiency, long span structures should have their shapes approximate thecoresponding moment diagram, hence an arch, suspended cable, or tendon configuration in a prestressedconcrete beam all are nearly parabolic, Fig. 20.1.

7 Long span structures can be built using flat construction such as girders or trusses. However, for spansin excess of 100 ft, it is often more economical to build a curved structure such as an arch, suspendedcable or thin shells.

8 Since the dawn of history, mankind has tried to span distances using arch construction. Essentiallythis was because an arch required materials to resist compression only (such as stone, masonary, bricks),and labour was not an issue.

9 The basic issues of static in arch design are illustrated in Fig. 20.2 where the vertical load is per unithorizontal projection (such as an external load but not a self-weight). Due to symmetry, the verticalreaction is simply V = wL

2 , and there is no shear across the midspan of the arch (nor a moment). Takingmoment about the crown,

M = Hh− wL2

(L

2− L

4

)= 0 (20.1)

Solving for H

H =wL2

8h(20.2)

We recall that a similar equation was derived for arches., and H is analogous to the C − T forces in abeam, and h is the overall height of the arch, Since h is much larger than d, H will be much smaller


M = w L /82

L

w=W/L

BEAMBEAM

W/2 W/2

- C

+ T

- C

+ T

RISE = h

SAG = h

IDEALISTIC ARCHSHAPE GIVEN BYMOMENT DIAGRAM

IDEALISTIC SUSPENSIONSHAPE GIVEN BYMOMENT DIAGRAM

T

C

T

C

NOTE THAT THE "IDEAL" SHAPE FOR AN ARCH OR SUSPENSION

SYSTEM IS EQUIVILENT TO THE DESIGN LOAD MOMENT DIAGRAM

M-ARM smallC-T large

Figure 20.1: Moment Resisting Forces in an Arch or Suspension System as Compared to a Beam, (Linand Stotesbury 1981)

CROWN2

M = VL/2 - wL /8 - H h = 0

BASE2M = wL /8 - H h = 0

w

V = wL/2

H

w

h

L/2

h

V = wL/2

H H

VR R

L

22R = V + H

H = wL /8h2

wL/2

Figure 20.2: Statics of a Three-Hinged Arch, (Lin and Stotesbury 1981)



than C − T in a beam.

10 Since equilibrium requiresH to remain constant across thee arch, a parabolic curve would theoreticallyresult in no moment on the arch section.

11 Three-hinged arches are statically determinate structures which shape can acomodate support settle-ments and thermal expansion without secondary internal stresses. They are also easy to analyse throughstatics.

12 An arch carries the vertical load across the span through a combination of axial forces and flexuralones. A well dimensioned arch will have a small to negligible moment, and relatively high normalcompressive stresses.

13 An arch is far more efficient than a beam, and possibly more economical and aesthetic than a trussin carrying loads over long spans.

14 If the arch has only two hinges, Fig. 20.3, or if it has no hinges, then bending moments may existeither at the crown or at the supports or at both places.


h

h’

V V

M

w

h

H’=wl /8h’<2

wl /8h2

H’H’<H


V V

H

M

w

h

L

crownM

M

base

baseh’

H’<H

Figure 20.3: Two Hinged Arch, (Lin and Stotesbury 1981)

15 Since H varies inversely to the rise h, it is obvious that one should use as high a rise as possible. Fora combination of aesthetic and practical considerations, a span/rise ratio ranging from 5 to 8 or perhapsas much as 12, is frequently used. However, as the ratio goes higher, we may have buckling problems,and the section would then have a higher section depth, and the arch advantage diminishes.

16 In a parabolic arch subjected to a uniform horizontal load there is no moment. However, in practicean arch is not subjected to uniform horizontal load. First, the depth (and thus the weight) of an archis not usually constant, then due to the inclination of the arch the actual self weight is not constant.Finally, live loads may act on portion of the arch, thus the line of action will not necessarily follow thearch centroid. This last effect can be neglected if the live load is small in comparison with the dead load.

17 Since the greatest total force in the arch is at the support, (R =√V 2 +H2), whereas at the crown

we simply have H , the crown will require a smaller section than the support.


h

h’

V V

M

w

h

H’=wl /8h’<2

wl /8h2

H’H’<H


V V

H

M

w

h

L

crownM

M

base

baseh’

H’<H

Figure 20.4: Arch Rib Stiffened with Girder or Truss, (Lin and Stotesbury 1981)



20.1.1 Statically Determinate

Example 20-1: Three Hinged Arch, Point Loads. (Gerstle 1974)

Determine the reactions of the three-hinged arch shown in Fig. 20.5

Figure 20.5:

Solution:

Four unknowns, three equations of equilibrium, one equation of condition ⇒ statically determinate.

(+ �✁✛) ΣMC

z = 0; (RAy)(140) + (80)(3.75)− (30)(80)− (20)(40) +RAx(26.25) = 0⇒ 140RAy + 26.25RAx = 2.900

(+✲ )ΣFx = 0; 80−RAx −RCx = 0(+ ✻)ΣFy = 0; RAy +RCy − 30− 20 = 0

(+ �✁✛)ΣMB

z = 0; (Rax)(60)− (80)(30)− (30)(20) + (RAy)(80) = 0⇒ 80RAy + 60RAx = 3, 000

(20.3)Solving those four equations simultaneously we have:

140 26.25 0 00 1 0 11 0 1 080 60 0 0

RAyRAxRCyRCx

=

2, 9008050

3, 000

⇒

RAyRAxRCyRCx

=

15.1 k

29.8 k

34.9 k

50.2 k

(20.4)

We can check our results by considering the summation with respect to b from the right:

(+ �✁✛)ΣMB

z = 0;−(20)(20)− (50.2)(33.75) + (34.9)(60) = 0√

(20.5)

Example 20-2: Semi-Circular Arch, (Gerstle 1974)

Determine the reactions of the three hinged statically determined semi-circular arch under its owndead weight w (per unit arc length s, where ds = rdθ). 20.6Solution:

I Reactions The reactions can be determined by integrating the load over the entire structure



R cosθ

R

A

B

C

R

A

B

θ

dP=wRdθ

θ

rθ

Figure 20.6: Semi-Circular three hinged arch

1. Vertical Reaction is determined first:

(+ �✁✛) ΣMA = 0;−(Cy)(2R) +

∫ θ=π

θ=0

wRdθ︸︷︷︸dP

R(1 + cos θ)︸︷︷︸moment arm

= 0 (20.6-a)

⇒ Cy =wR

2

∫ θ=π

θ=0

(1 + cos θ)dθ =wR

2[θ − sin θ] |θ=πθ=0

=wR

2[(π − sinπ)− (0− sin 0)]

= π2wR (20.6-b)

2. Horizontal Reactions are determined next

(+ �✁✛)ΣMB = 0;−(Cx)(R) + (Cy)(R)−

∫ θ=π2

θ=0

wRdθ︸︷︷︸dP

R cos θ︸︷︷︸moment arm

= 0 (20.7-a)

⇒ Cx =π

2wR − wR

2

∫ θ=π2

θ=0

cos θdθ =π

2wR − wR[sin θ] |θ=π

2θ=0 =

π

2wR − wR(

π

2− 0)

=(π2 − 1

)wR (20.7-b)

By symmetry the reactions at A are equal to those at C

II Internal Forces can now be determined, Fig. 20.7.

1. Shear Forces: Considering the free body diagram of the arch, and summing the forces in theradial direction (ΣFR = 0):

−(π

2− 1)wR︸︷︷︸Cx

cos θ +π

2wR︸︷︷︸Cy

sin θ −∫ θ

α=0

wRdα sin θ + V = 0 (20.8)

⇒ V = wR[(π2 − 1) cos θ + (θ − π

2 ) sin θ]

(20.9)

2. Axial Forces: Similarly, if we consider the summation of forces in the axial direction (ΣFN =0):

(π2 − 1)wR sin θ + π2wR cos θ −

∫ θ

α=0

wRdα cos θ +N = 0 (20.10)

⇒ N = wR[(θ − π

2 ) cos θ − (π2 − 1) sin θ]

(20.11)

3. Moment: Now we can consider the third equation of equilibrium (ΣMz = 0):

(+ �✁✛)ΣM (π2 − 1)wR ·R sin θ − π

2wR2(1− cos θ) +∫ θ

α=0

wRdα ·R(cosα− cos θ) +M = 0 (20.12)

⇒ M = wR2[π2 (1− sin θ) + (θ − π

2 ) cos θ]

(20.13)



R cos θ

R cos αR

cos

(

)

−αθθR(1-cos )

θR

sin

C =( /2-1)wRπx

C =

/2

wR

πy

dP=wRd

VN

M

θ α

d α

α

rθ

Figure 20.7: Semi-Circular three hinged arch; Free body diagram

III Deflection are determined last

1. The real curvature φ is obtained by dividing the moment by EI

φ =M

EI=wR2

EI

[π2(1 − sin θ) + (θ − π

2) cos θ

](20.14)

1. The virtual force δP will be a unit vertical point in the direction of the desired deflection,causing a virtual internal moment

δM =R

2[1− cos θ − sin θ] 0 ≤ θ ≤ π

2(20.15)

p2. Hence, application of the virtual work equation yields:

1︸︷︷︸δP

·∆ = 2∫ π

2

θ=0

wR2

EI

[π2(1− sin θ) + (θ − π

2) cos θ

]︸︷︷︸

MEI=φ

· R2· [1− cos θ − sin θ]︸︷︷︸

δM

Rdθ︸︷︷︸dx

=wR4

16EI[7π2 − 18π − 12

]= .0337wR

4

EI (20.16-a)

20.1.2 Statically Indeterminate

Example 20-3: Statically Indeterminate Arch, (Kinney 1957)



Determine the value of the horizontal reaction component of the indicated two-hinged solid rib arch,Fig. 20.8 as caused by a concentrated vertical load of 10 k at the center line of the span. Considershearing, axial, and flexural strains. Assume that the rib is a W24x130 with a total area of 38.21 in2,that it has a web area of 13.70 in2, a moment of inertia equal to 4,000 in4, E of 30,000 k/in2, and ashearing modulus G of 13,000 k/in2.

Figure 20.8: Statically Indeterminate Arch

Solution:

1. Consider that end C is placed on rollers, as shown in Fig. ?? A unit fictitious horizontal force isapplied at C. The axial and shearing components of this fictitious force and of the vertical reactionat C, acting on any section θ in the right half of the rib, are shown at the right end of the rib inFig. 13-7.

2. The expression for the horizontal displacement of C is

1︸︷︷︸δP

∆Ch = 2∫ B

C

δMM

EIds+ 2

∫ B

C

δVV

AwGds+ 2

∫ B

C

δNN

AEds (20.17)

3. From Fig. 20.9, for the rib from C to B,

M =P

2(100−R cos θ) (20.18-a)

δM = 1(R sin θ − 125.36) (20.18-b)

V =P

2sin θ (20.18-c)

δV = cos θ (20.18-d)

N =P

2cos θ (20.18-e)



Figure 20.9: Statically Indeterminate Arch; ‘Horizontal Reaction Removed

δN = − sin θ (20.18-f)ds = Rdθ (20.18-g)

4. If the above values are substituted in Eq. 20.17 and integrated between the limits of 0.898 andπ/2, the result will be

∆Ch = 22.55 + 0.023− 0.003 = 22.57 (20.19)

5. The load P is now assumed to be removed from the rib, and a real horizontal force of 1 k is assumedto act toward the right at C in conjunction with the fictitious horizontal force of 1 k acting to theright at the same point. The horizontal displacement of C will be given by

δChCh = 2∫ B

C

δMM

EIds+ 2

∫ B

C

δVV

AwGds+ 2

∫ B

C

δNN

AEds (20.20-a)

= 2.309 + 0.002 + 0.002 = 2.313 in (20.20-b)

6. The value of the horizontal reaction component will be

HC =∆Ch

δChCh=

22.572.313

= 9.75 k (20.21)

7. If only flexural strains are considered, the result would be

HC =22.552.309

= 9.76 k (20.22)

Comments

1. For the given rib and the single concentrated load at the center of the span it is obvious that theeffects of shearing and axial strains are insignificant and can be disregarded.

2. Erroneous conclusions as to the relative importance of shearing and axial strains in the usual solidrib may be drawn, however, from the values shown in Eq. 20.19. These indicate that the effects ofthe shearing strains are much more significant than those of the axial strains. This is actually thecase for the single concentrated load chosen for the demonstration, but only because the rib doesnot approximate the funicular polygon for the single load. As a result, the shearing componentson most sections of the rib are more important than would otherwise be the case.



3. The usual arch encountered in practice, however, is subjected to a series of loads, and the axis ofthe rib will approximate the funicular polygon for these loads. In other words, the line of pressureis nearly perpendicular to the right section at all points along the rib. Consequently, the shearingcomponents are so small that the shearing strains are insignificant and are neglected.

4. Axial strains, resulting in rib shortening, become increasingly important as the rise-to-span ratioof the arch decreases. It is advisable to determine the effects of rib shortening in the design ofarches. The usual procedure is to first design the rib by considering flexural strains only, and thento check for the effects of rib shortening.

20.2 Curved Space Structures

Example 20-4: Semi-Circular Box Girder, (Gerstle 1974)

Determine the reactions of the semi-circular cantilevered box girder shown in Fig. 20.10 subjected

αd

A

y

x

z

O

R

A

B

C

y

x

z

θ

θ

rΤ ΖΜ

V

wRdα

α

Figure 20.10: Semi-Circular Box Girder

to its own weight w.Solution:

I Reactions are again determined first From geometry we have OA = R, OB = R cos θ, AB = OA −BO = R − R cos θ, and BP = R sin θ. The moment arms for the moments with respect to the xand y axis are BP and AB respectively. Applying three equations of equilibrium we obtain

FAz −∫ θ=π

θ=0

wRdθ = 0 ⇒ FAz = wRπ (20.23-a)

MAx −

∫ θ=π

θ=0

(wRdθ)(R sin θ) = 0 ⇒ MAx = 2wR2 (20.23-b)

MAy −

∫ θ=π

θ=0

(wRdθ)R(1 − cos θ) = 0 ⇒ MAy = −wR2π (20.23-c)

II Internal Forces are determined next

1. Shear Force:

(+ ✻)ΣFz = 0⇒ V −∫ θ

0

wRdα = 0⇒ V = wrθ (20.24)



2. Bending Moment:

ΣMR = 0⇒M −∫ θ

0

(wRdα)(R sinα) = 0⇒ M = wR2(1 − cos θ) (20.25)

3. Torsion:

ΣMT = 0⇒ +∫ θ

0

(wRdα)R(1 − cosα) = 0⇒ T = −wR2(θ − sin θ) (20.26)

III Deflection are determined last we assume a rectangular cross-section of width b and height d = 2band a Poisson’s ratio ν = 0.3.

1. Noting that the member will be subjected to both flexural and torsional deformations, weseek to determine the two stiffnesses.

2. The flexural stiffness EI is given by EI = E bd3

12 = E b(2b)3

12 = 2Eb4

3 = .667Eb4.

3. The torsional stiffness of solid rectangular sections J = kb3d where b is the shorter side of thesection, d the longer, and k a factor equal to .229 for d

b = 2. Hence G = E2(1+ν) = E

2(1+.3) =.385E, and GJ = (.385E)(.229b4) = .176Eb4.

4. Considering both flexural and torsional deformations, and replacing dx by rdθ:

δP∆︸︷︷︸δW

∗

=∫ π

0

δMM

EIzRdθ︸︷︷︸

Flexure

+∫ π

0

δTT

GJRdθ︸︷︷︸

Torsion︸︷︷︸δU

∗

(20.27)

where the real moments were given above.

5. Assuming a unit virtual downward force δP = 1, we have

δM = R sin θ (20.28-a)δT = −R(1− cos θ) (20.28-b)

6. Substituting these expression into Eq. 20.27

1︸︷︷︸δP

∆ =wR2

EI

∫ π

0

(R sin θ)︸︷︷︸M

(1− cos θ)︸︷︷︸δM

Rdθ +wR2

GJ

∫ π

0

(θ − sin θ)︸︷︷︸δT

R(1− cos θ)︸︷︷︸T

Rdθ

=wR4

EI

∫ π

0

[(sin θ − sin θ cos θ) +

1.265

(θ − θ cos θ − sin θ + sin θ cos θ)]dθ

=wR4

EI( 2.︸︷︷︸Flexure

+ 18.56︸︷︷︸Torsion

)

= 20.56wR4

EI (20.29-a)

20.2.1 Theory

ßAdapted from (Gerstle 1974)

18 Because space structures may have complicated geometry, we must resort to vector analysis1 todetermine the internal forces.

19 In general we have six internal forces (forces and moments) acting at any section.1To which you have already been exposed at an early stage, yet have very seldom used it so far in mechanics!



20.2.1.1 Geometry

20 In general, the geometry of the structure is most conveniently described by a parameteric set ofequations

x = f1(θ); y = f2(θ); z = f3(θ) (20.30)

as shown in Fig. 20.11. the global coordinate system is denoted by X − Y −Z, and its unit vectors are

Figure 20.11: Geometry of Curved Structure in Space

denoted2 i, j,k.

21 The section on which the internal forces are required is cut and the principal axes are identified asN − S −W which correspond to the normal force, and bending axes with respect to the Strong andWeak axes. The corresponding unit vectors are n, s,w.

22 The unit normal vector at any section is given by

n =dxi + dyj+ dzk

ds=

dxi + dyj+ dzk(dx2 + dy2 + dz2)1/2

(20.31)

23 The principal bending axes must be defined, that is if the strong bending axis is parallel to the XYplane, or horizontal (as is generally the case for gravity load), then this axis is normal to both the Nand Z axes, and its unit vector is

s =n×k|n×k| (20.32)

24 The weak bending axis is normal to both N and S, and thus its unit vector is determined from

w = n×s (20.33)

20.2.1.2 Equilibrium

25 For the equilibrium equations, we consider the free body diagram of Fig. 20.12 an applied load P2All vectorial quantities are denoted by a bold faced character.



Figure 20.12: Free Body Diagram of a Curved Structure in Space

is acting at point A. The resultant force vector F and resultant moment vector M acting on the cutsection B are determined from equilibrium

ΣF = 0; P+ F = 0; F = −P (20.34-a)ΣMB = 0; L×P+M = 0; M = −L×P (20.34-b)

where L is the lever arm vector from B to A.

26 The axial and shear forces N,Vs and Vw are all three components of the force vector F along theN,S, and W axes and can be found by dot product with the appropriate unit vectors:

N = F·n (20.35-a)Vs = F·s (20.35-b)Vw = F·w (20.35-c)

27 Similarly the torsional and bending moments T,Ms and Mw are also components of the momentvector M and are determined from

T = M·n (20.36-a)Ms = M·s (20.36-b)Mw = M·w (20.36-c)

28 Hence, we do have a mean to determine the internal forces. In case of applied loads we summ, andfor distributed load we integrate.

Example 20-5: Internal Forces in an Helicoidal Cantilevered Girder, (Gerstle 1974)

Determine the internal forces N,Vs, and Vw and the internal moments T,Ms and Mw along thehelicoidal cantilevered girder shown in FIg. 20.13 due to a vertical load P at its free end.Solution:

1. We first determine the geometry in terms of the angle θ

x = R cos θ; y = R sin θ; z =H

πθ (20.37)



Figure 20.13: Helicoidal Cantilevered Girder



2. To determine the unit vector n at any point we need the derivatives:

dx = −R sin θdθ; dy = R cos θdθ; dz =H

πdθ (20.38)

and then insert into Eq. 20.31

n =−R sin θi+R cos θj+H/πk[

R2 sin2 θ +R2 cos2 θ + (H/π)2]1/2 (20.39-a)

=1[

1 + (H/πR)2]1/2︸︷︷︸

K

[sin θi+ cos θj+ (H/πR)k] (20.39-b)

Since the denominator depends only on the geometry, it will be designated by K.

3. The strong bending axis lies in a horizontal plane, and its unit vector can thus be determined fromEq. 20.32:

n×k =1K

∣∣∣∣∣∣i j k

− sin θ cos θ HπR

0 0 1

∣∣∣∣∣∣ (20.40-a)

=1K

(cos θi+ sin θj) (20.40-b)

and the absolute magnitude of this vector |k×n| = 1K , and thus

s = cos θi+ sin θj (20.41)

4. The unit vector along the weak axis is determined from Eq. 20.33

w = s×n =1K

∣∣∣∣∣∣i j k

cos θ sin θ 0− sin θ cos θ H

πR

∣∣∣∣∣∣ (20.42-a)

=1K

(H

πRsin θi− H

πRcos θj+ k

)(20.42-b)

5. With the geometry definition completed, we now examine the equilibrium equations. Eq. 20.34-aand 20.34-b.

ΣF = 0; F = −P (20.43-a)ΣMb = 0; M = −L×P (20.43-b)

where

L = (R−R cos θ)i+ (0 −R sin θ)j+(

0− θπH

)k (20.44)

and

L×P = R

∣∣∣∣∣∣i j k

(1− cos θ) − sin θ − θπHR

0 0 P

∣∣∣∣∣∣ (20.45-a)

= PR[− sin θi− (1− cos θ)j] (20.45-b)

andM = PR[sin θi + (1− cos θ)j] (20.46)



6. Finally, the components of the force F = −Pk and the moment M are obtained by appropriatedot products with the unit vectors

N = F·n = − 1KP

HπR (20.47-a)

Vs = F·s = 0 (20.47-b)

Vw = F·w = − 1KP (20.47-c)

T = M·n = −PRK (1− cos θ) (20.47-d)

Ms = M·s = PR sin θ (20.47-e)

Mw = M·w = PHπK (1− cos θ) (20.47-f)


Draft

Chapter 21

STATIC INDETERMINANCY;FLEXIBILITY METHOD

All the examples in this chapter are taken verbatim from White, Gergely and Sexmith

21.1 Introduction

1 A statically indeterminate structure has more unknowns than equations of equilibrium (and equationsof conditions if applicable).

2 The advantages of a statically indeterminate structures are:

1. Lower internal forces

2. Safety in redundancy, i.e. if a support or members fails, the structure can redistribute its internalforces to accomodate the changing B.C. without resulting in a sudden failure.

3 Only disadvantage is that it is more complicated to analyse.

4 Analysis mehtods of statically indeterminate structures must satisfy three requirements

Equilibrium

Force-displacement (or stress-strain) relations (linear elastic in this course).

Compatibility of displacements (i.e. no discontinuity)

5 This can be achieved through two classes of solution

Force or Flexibility method;

Displacement or Stiffness method

6 The flexibility method is first illustrated by the following problem of a statically indeterminate cablestructure in which a rigid plate is supported by two aluminum cables and a steel one. We seek todetermine the force in each cable, Fig. 21.1

1. We have three unknowns and only two independent equations of equilibrium. Hence the problemis statically indeterminate to the first degree.

2. Applying the equations of equilibrium

ΣMz = 0; ⇒ P leftAl = P right

Al

ΣFy = 0; ⇒ 2PAl + PSt = P (21.1-a)

Thus we effectively have two unknowns and one equation.

Draft21–2 STATIC INDETERMINANCY; FLEXIBILITY METHOD

Figure 21.1: Statically Indeterminate 3 Cable Structure

3. We need to have a third equation to solve for the three unknowns. This will be derived from thecompatibility of the displacements in all three cables, i.e. all three displacements must beequal:

DAl = DSt (21.2)

4. Finally, those isplacements are obtained from the Force-Displacement relations:

σ = PA

ε = ∆LL

ε = σE

⇒ ∆L =

PL

AE(21.3)

PAlL

EAlAAl︸︷︷︸DAl

=PStL

EStASt︸︷︷︸DSt

⇒ PAlPSt

=(EA)Al(EA)St

(21.4)

or−(EA)StPAl + (EA)AlPSt = 0 (21.5)

5. Solution of Eq. 21.1-a and 21.5 yields[2 1

−(EA)St (EA)Al

]{PAlPSt

}={P0

}⇒{PAlPSt

}=[

2 1−(EA)St (EA)Al

]−1{P0

}=

12(EA)Al + (EA)St︸︷︷︸

Determinant

[(EA)Al −1(EA)St 2

]{P0

}

6. We observe that the solution of this sproblem, contrarily to statically determinate ones, dependson the elastic properties.

7 Another example is the propped cantiliver beam of length L, Fig. 21.2

1. First we remove the roller support, and are left with the cantilever as a primary structure.

2. We then determine the deflection at point B due to the applied load P using the virtual forcemethod

1.D =∫δM

M

EIdx (21.6-a)

=∫ L/2

0

0−pxEI

dx+∫ L/2

0

−(PL

2+ Px

)(−x)dx (21.6-b)


Draft21.1 Introduction 21–3

x x

fBB

P

B AC L/2 L/2

P

D

1

QL/2

PL/4

-

+-

-

-PL PL/2

-(1)

L/2

Primary Structure Under Actual Load

Primary Structure Under Redundant Loading

Bending Moment Diagram

Figure 21.2: Propped Cantilever Beam



=1EI

∫ L/2

0

(PL

2x+ Px2

)dx (21.6-c)

=1EI

[PLx2

4+Px3

3

]∣∣∣∣L/20

(21.6-d)

=548PL3

EI(21.6-e)

3. We then apply a unit load at point B and solve for the displacement at B using the virtual forcemethod

1fBB =∫δM

M

EIdx (21.7-a)

=∫ L/2

0

(x)x

EIdx (21.7-b)

=(1)L3

24EI(21.7-c)

4. Then we argue that the displacement at point B is zero, and hence the displacement fBB shouldbe multiplied by RB such that

RBfBB = D (21.8)

to ensure compatibility of displacements, hence

RB =D

fBB=

548PL

3EI(1)L3

24EI

= 52P (21.9)

21.2 The Force/Flexibility Method

8 Based on the previous two illustrative examples, we now seek to develop a general method for thelinear elastic analysis of statically indeterminate structures.

1. Identify the degree of static indeterminancy (exterior and/or interior) n.

2. Select n redundant unknown forces and/or couples in the loaded structure along with n corre-sponding releases (angular or translation).

3. The n releases render the structure statically determinate, and it is called the primary structure.

4. Determine the n displacements in the primary structure (with the load applied) corresponding tothe releases, Di.

5. Apply a unit force at each of the releases j on the primary structure, without the external load, anddetermine the displacements in all releases i, we shall refer to these displacements as the flexibilitycoefficients, fij , i.e. displacement at release i due to a unit force at j

6. Write the compatibility of displacement relationf11 f12 · · · f1nf21 f22 · · · f2n· · · · · · · · · · · ·fn1 fn2 · · · fnn

︸︷︷︸[f ]

R1

R2

· · ·Rn

︸︷︷︸

R

D1

D2

· · ·Dn

︸︷︷︸

D

=

D01

D02

· · ·D0n

︸︷︷︸

D0i

(21.10)

and

[R] = [f ]−1{D−D0

i

}(21.11)

Note that D0i is the vector of initial displacements, which is usually zero unless we have an initial

displacement of the support (such as support settlement).


Draft21.3 Short-Cut for Displacement Evaluation 21–5

7. The reactions are then obtained by simply inverting the flexibility matrix.

9 Note that from Maxwell-Betti’s reciprocal theorem, the flexibility matrix [f ] is always symmetric.

21.3 Short-Cut for Displacement Evaluation

10 Since deflections due to flexural effects must be determined numerous times in the flexibility method,Table 21.1 may simplify some of the evaluation of the internal strain energy. You are strongly discouragedto use this table while still at school!.

g2(x)

g1(x) L

a ❍❍❍

L

a✥✥✥

La b

L

c

Lac Lac2

Lc(a+b)2

❍❍❍

L

cLac2

Lac3

Lc(2a+b)6

✟✟✟

L

cLac2

Lac6

Lc(a+2b)6

✥✥✥

Lc d

La(c+d)2

La(2c+d)6

La(2c+d)+Lb(c+2d)6

Lc d e

La(c+4d+e)6

La(c+2d)6

La(c+2d)+Lb(2d+e)6

Table 21.1: Table of∫ L

0

g1(x)g2(x)dx

21.4 Examples

Example 21-1: Steel Building Frame Analysis, (White et al. 1976)

A small, mass-produced industrial building, Fig. 21.3, is to be framed in structural steel with atypical cross section as shown below. The engineer is considering three different designs for the frame:(a) for poor or unknown soil conditions, the foundations for the frame may not be able to develop anydependable horizontal forces at its bases. In this case the idealized base conditions are a hinge at one ofthe bases and a roller at the other; (b) for excellent soil conditions with properly designed foundations,the bases of the frame legs will have no tendency to move horizontally, and the idealized base conditionis that of hinges at both points A and D; and (c) a design intermediate to the above cases, with a steeltie member capable of carrying only tension running between points A and D in the floor of the building.The foundations would not be expected to provide any horizontal restraint for this latter case, and thehinge-roller details at points A and D would apply.

Critical design loads for a frame of this type are usually the gravity loads (dead load + snow load)and the combination of dead load and wind load. We will restrict our attention to the first combination,and will use a snow load of 30 psf and an estimated total dead load of 20 psf. With frames spaced at 15ft on centers along the length of the building, the design load is 15 (30 + 20) = 750 lb/ft.



Figure 21.3:

If the frame is made of steel beam sections 21 in. deep and weighing 62 lb/ft of length (W 21 × 62),and the tie member for design (c) is tentatively chosen as a 2-in.2 bar, determine the bending momentdiagrams for the three designs and discuss the alternate solutions.Solution:

Structure a This frame is statically determinate since it has three possible unknown externalforces acting on it, and the bending moment is shown in Fig. 21.6-a.

Structure b Hinging both legs of the frame results in another unknown force, making the structurestatically indeterminate to the first degree (one redundant).1. A lateral release at point A is chosen. with the redundant shearing force R1. The displacement D1Q

in the primary structure, as a result of the real loading, is shown in Figure 21.4-a. D1Q is computed byvirtual work.2. The virtual force system in Figure 21.4-b produces a virtual bending moment δM , which is uniform

Figure 21.4:

across the top member of the frame. The virtual moment acting through the real angle changes givesthe internal work term ∫

δMdφ =∫ 40

0

δMM

EIdx (21.12)



3. Equating this to the external virtual work of 1 ·D1Q, we have

1 ·D1Q =∫ 40

0

(12)(1/2)(.75)(Lx− x2)EI

dx (21.13)

orD1Q =

48, 000EI

k ft3 → (21.14)

4. The equation of consistent displacement is D1Q+f11R1 = 0. The flexibility coefficient f11 is computedby applying a unit horizontal force at the release and determining the displacement at the same point.5. It is seen that the real loading and the virtual loading are identical for this calculation, and

1 · f11 = 2[∫ 12

0

x2dx

EI+∫ 20

0

122dxEI

](21.15)

orf11 =

6, 912EI

k ft3 → (21.16)

6. Solving for R11EI

[48, 000 + 6, 912R1] = 0 (21.17)

orR1 = 6.93 k← (21.18)

7. The bending moment diagram is given below

Structure c The frame with the horizontal tie between points A and D has three unknown externalforces. However, the structure is statically indeterminate to the first degree since the tie member providesone degree of internal redundancy.8. The logical release to choose is a longitudinal release in the tie member, with its associated longitudinaldisplacement and axial force.9. The primary structure Fig. 21.5 is the frame with the tie member released. The compatibility

Figure 21.5:

equation is based on the fact that the displacement at the release must be zero; that is, the relativedisplacement of the two sections of the tie at the point of release must be zero, or

D1Q + f11R1 = 0 (21.19)



whereD1Q = displacement at release 1 in the primary structure, produced by the loading,f11 = relative displacement at release 1 for a unit axial force in the tie member,R1 = force in the tie member in the original structure.

10. Virtual work is used to determine both displacement terms.11. The value of D1Q is identical to the displacement D1Q computed for structure (b) because the tiemember has no forces (and consequently no deformations) in the primary structure. Thus

D1Q =(48, 000)(1, 728)(30 · 103)(1, 327) = 2.08 in✲ (21.20)

12. The flexibility coefficient f11 is composed of two separate effects: a flexural displacement due to theflexibility of the frame, and the axial displacement of the stressed tie member. The virtual and realloadings for this calculation are shown in Figure 21.5. The virtual work equation is

1 · f11 = 2[∫ 12

0

x21dx1EI

+∫ 20

0

(12)2dxEI

]+ δP

pL

EA(21.21-a)

=6, 912EI

+1(1)(40)EA

(21.21-b)

=(6, 912)(1, 728)

(30 · 103)(1, 327) +40(12)

(30 · 103)(2) (21.21-c)

= 0.300 + 0.008 = 0.308 (21.21-d)

andf11 = 0.308in./k (21.22)

13. The equation of consistent deformation is

D1Q + f11R1 = 0 (21.23)

or2.08− .308R1 = 0 (21.24)

orR1 = 6.75 k (tension) (21.25)

14. The two displacement terms in the equation must carry opposite signs to account for their differencein direction.

Comments The bending moment in the frame, Figure 21.6-c differs only slightly from that ofstructure (b). In other words, the tie member has such high axial stiffness that it provides nearly asmuch restraint as the foundation of structure (b). Frames with tie members are used widely in industrialbuildings. A lesson to be learned here is that it is easy to provide high stiffness through an axially loadedmember.

Figure 21.6:

The maximum moment in frames (b) and (c) is about 55% of the maximum moment in frame (a).This effect of continuity and redundancy is typical – the positive bending moments in the members arelowered while the joint moments increase and a more economical design can be realized.

Finally, we should notice that the vertical reactions at the bases of the columns do not change withthe degree of horizontal restraint at the bases. A question to ponder is “Does this type of reactionbehavior occur in all frames, or only in certain geometrical configurations?”



Example 21-2: Analysis of Irregular Building Frame, (White et al. 1976)

The structural steel frame for the Church of the Holy Spirit, Penfield, New York is shown below. Inthis example we will discuss the idealization of the structure and then determine the forces and bendingmoments acting on the frame.Solution:

1. A sectional view of the building is given in Figure 21.8-a.

Figure 21.7:

The two main horizontal members of the frame are supported at points A and D by masonry walls.2. The connection used at these points is not intended to transmit axial forces from the frame to thewall; accordingly, the axial forces in the horizontal members are assumed to be zero and the joints at Aand D are idealized as rollers that transmit vertical forces only.3. The base joint E is designed to resist both horizontal and vertical loads, but not moment, and isassumed to be a hinge.4. Finally, joints B and C are designed to provide continuity and will be taken as rigid; that is, theangles of intersection of the members at the joint do not change with applied loading.5. The frame is simplified for analysis by removing the small 4-in. wide flange members EF and FGand replacing their load effect by applying the roof load which acts on EF directly to the segment AG.6. The idealized frame is shown in Figure 21.8-b.7. The dead load on the higher portion of the frame is wAB = 25 psf times the frame spacing of 13.33ft, or wAB = 25(13.33) = 334 lb/ft along the frame.8. The dead load on CD is less because the weight of the frame member is substantially smaller, andthe dead load is about 19 psf, or wCD = 19(13.33) = 254 lb/ft of frame.9. Snow load is 35 psf over both areas, or w = 35(13.33) = 467 lb/ft.10. Total loads are then



Figure 21.8:



Member AB: w = 334 + 467 = 800 lb/ft = 0.8 k/ftMember CD: w = 254 + 467 = 720 lb/ft = 0.72 k/ft

11. The frame has four unknown reaction components and therefore has one redundant. Althoughseveral different releases are possible, we choose an angular (bending) release at point B.12. The resulting primary structure is shown in Figure 21.8-c, where the redundant quantity R1 is thebending moment at point B.13. The equation of compatibility is

θ1Q + θ11R1 = 0 (21.26)

where θ1Q is the relative angular rotation corresponding to release 1 as produced by the actual loading,and θ11 is the flexibility coefficient for a unit moment acting at the release.14. From virtual work we have

1 · θ1Q =∫δMdφ =

∫δM

M

EIdx (21.27)

and1 · θ11 =

∫δMdφ =

∫δM

M

EIdx (21.28)

where m is a real unit load and δM and M are defined in Figure 21.8-d and e.15. Then

θ1Q =1

(EI)AB

∫ 61.42

0

x

61.42(24.6x− 0.4x2)dx (21.29-a)

+1

(EI)CD

∫ 21.33

0

x

21.33(7.68x− 0.36x2)dx (21.29-b)

=7750

(EI)AB+

295(EI)CD

(21.29-c)

16. with IAB = 4, 470 in.4 and ICD = 290 in.4

θ1Q =1.733E

+1.0107E

=2.75E

(21.30)

17. Similarly

θ11 =1

(EI)AB

∫ 61.42

0

( x

61.42

)2dx (21.31-a)

+1

(EI)CD

∫ 21.33

0

( x

21.33

)2dx+

1(EI)BC

∫ 8

0

(1)2dx (21.31-b)

18. with IBC = 273 in4

θ11 =20.5

(EI)AB+

7.11(EI)CD

+8.0

(EI)BC=

0.0585E

(21.32)

Note that the numerators of θ1Q and θ11 have the units k-ft2/in4.19. Applying the compatibility equation,

2.75E

+0.0585E

R1 = 0 (21.33)

and the bending moment at point B is R1 = −47.0 ft-k . The reactions and moments in the structureare given in Figure 21.8-f.

Example 21-3: Redundant Truss Analysis, (White et al. 1976)



Figure 21.9:

Determine the bar forces in the steel truss shown below using the force method of analysis. The trussis part of a supporting tower for a tank, and the 20 kN horizontal load is produced by wind loading onthe tank.Solution:

1. Applying the criteria for indeterminacy, 2× 4 = 8 equations, 6 members + 3 reactions ⇒ one degreeof indeterminacy. A longitudinal release in any of the six bars may be chosen.2. Because the truss members carry only axial load, a longitudinal release is identical to actually cuttingthe member and removing its axial force capability from the truss.3. In analyzing trusses with double diagonals it is both convenient and customary to select the releasein one of the diagonal members because the resulting primary structure will be the conventional trussform to which we are accustomed.4. Choosing the diagonal member BC for release, we cut it and remove its axial stiffness from thestructure. The primary structure is shown in Figure 21.9-b.5. The analysis problem reduces to applying an equation of compatibility to the changes in length ofthe release member. The relative displacement D1Q of the two cut ends of member BC, as produced bythe real loading, is shown in Figure 21.9-c.6. The displacement is always measured along the length of the redundant member, and since theredundant is unstressed at this stage of the analysis, the displacement D1Q is equal to the relativedisplacement of joint B with respect to joint C.7. This displacement must be eliminated by the relative displacements of the cut ends of member BCwhen the redundant force is acting in the member. The latter displacement is written in terms of theaxial flexibility coefficient f11, and the desire equation of consistent deformation is

D1Q + f11R1 = 0 (21.34)

8. The quantity D1Q is given by

1 ·D1Q = ΣδP (∆l) = ΣδP (PL/AE) (21.35)



where δP and P are given in Figure 21.9-d and c, respectively.9. Similarly,

f11 = ΣδP (PL/AE) (21.36)

10. Evaluating these summations in tabular form:

Member P δP L δPPL δPPLAB 0 −0.707 3 0 1.5BC 0 −0.707 3 0 1.5CD +20 −0.707 3 −42.42 1.5AC +20 −0.707 3 −42.42 1.5AD −28.28 +1 4.242 −119.96 4.242BC 0 +1 4.242 0 4.242

-204.8 14.484

11. Since A = constant for each member

D1Q = ΣδPPL

AE= −−204.8

AEandf11 =

14.484AE

(21.37)

then1AE

[−204.8 + 14.484R1] = 0 (21.38)

12. The solution for the redundant force value is R1 = 14.14 kN .13. The final values for forces in each of the truss members are given by superimposing the forces dueto the redundant and the forces due to the real loading.14. The real loading forces are shown in Figure 21.9-c while the redundant force effect is computed bymultiplying the member forces in Figure 21.9-d by 2.83, the value of the redundant.15. It is informative to compare the member forces from this solution to the approximate analysisobtained by assuming that the double diagonals each carry half the total shear in the panel. Thecomparison is given in Figure 21.10; it reveals that the approximate analysis is the same as the exact

Figure 21.10:

analysis for this particular truss. The reason for this is that the stiffness provided by each of the diagonalmembers (against “shear” deformation of the rectangular panel) is the same, and therefore they eachcarry an equal portion of the total shear across the panel.16. How would this structure behave if the diagonal members were very slender?

Example 21-4: Truss with Two Redundants, (White et al. 1976)



Another panel with a second redundant member is added to the truss of Example 21-3 and the newtruss is supported at its outermost lower panel points, as shown in Figure 21.11. The truss, which issimilar in form to trusses used on many railway bridges, is to be analyzed for bar forces under the givenloading.Solution:

1. The twice redundant truss is converted to a determinate primary structure by releasing two membersof the truss; we choose two diagonals (DB and BF ).2. Releasing both diagonals in a single panel, such as members AE and DB, is inadmissible since itleads to an unstable truss form.3. The member forces and required displacements for the real loading and for the two redundant forces

Figure 21.11:

in members DB and BF are given in Figure 21.11.4. Although the real loading ordinarily stresses all members of the entire truss, we see that the unit forcescorresponding to the redundants stres only those members in the panel that contains the redundant; allother bar forces are zero.5. Recognizing this fact enables us to solve the double diagonal truss problem more rapidly than a framewith multiple redundants.6. The virtual work equations for computing the six required displacements (two due to load and fourflexibilities) are

1 ·D1Q = ΣδP 1

(PL

AE

)(21.39-a)

1 ·D2Q = ΣδP 2

(PL

AE

)(21.39-b)

1 · f11 = ΣδP 1

(P 1L

AE

)(21.39-c)

1 · f21 = ΣδP 2

(P 1L

AE

)(21.39-d)

f12 = f21 by the reciprocal theorem (21.39-e)



1 · f22 = ΣδP 2P 2L

AE(21.39-f)

7. If we assume tension in a truss member as postive, use tensile unit loads when computing the flexibilitycoefficients corresponding to the redundants, and let all displacement terms carry their own signs, thenin the solution for the redundants a positive value of force indicates tension while a negative value meansthe member is in compression.8. The calculation of f22 involves only the six members in the left panel of the truss; f21 involves onlymember BE.9. The simple procedures used for performing the displacement analyses, as summarized in tabular formin Table 21.2, leads one quickly to the compatibility equations which state that the cut ends of both

D1Q D2Q f11 f21 f22Member P P 1 P 2 L δP 1PL δP 2PL δP 1P 1L δP 2P 1L δP 2P 2LAB -9.5 -0.707 0 120 +806 0 60 0 0BC -9.5 0 -0.707 120 0 +806 0 0 60CF -9.5 0 -0.707 120 0 +806 0 0 60EF 0 0 -0.707 120 0 0 0 0 60DE +4 -0.707 0 120 -340 0 60 0 0AD -5.5 -0.707 0 120 +466 0 60 0 0AE +7.78 +1 0 170 +1,322 0 170 0 0BE .15 -0.707 -0.707 120 +1,272 +1272 60 60 60CE +13.43 0 +1 170 0 +2,280 0 0 170BD 0 +1 0 170 0 0 170 0 0BF 0 0 +1 170 0 0 0 0 170

+3,528 +5,164 +580 +60 +580

Table 21.2:

redundant members must match (there can be no gaps or overlaps of members in the actual structure).10. The equations are

D1Q + f11R1 + f12R2 = 0 (21.40-a)D2Q + f21R1 + f22R2 = 0 (21.40-b)

or1AE

[580 6060 580

] [R1

R2

]= − 1

AE

[3, 5285, 164

](21.41)

and

R1 = 5.20 k (21.42-a)

R2 = −8.38 k (21.42-b)

11. The final set of forces in the truss is obtained by adding up, for each member, the three separateeffects. In terms of the forces shown in Figure 21.11 and Table 21.2, the force in any member is given byF = P +R1P 1 + R2P 2. The final solution is given in Figure 21.11-e. The truss is part of a supportingtower for a tank, and the 20 kN horizontal load is produced by wind loading on the tank.12. NOTE: A mixture of internal redundant forces and external redundant reactions is no more difficultthan the preceding example. Consider the two-panel truss of this example modified by the addition ofanother reaction component at joint E, Figure 21.12. The three releases for this truss can be chosenfrom a number of possible combinations: diagonals DB and BF and the reaction at E; the same twodiagonals and the reaction at F ; the same diagonals and the top chord member BC, etc. The onlyrequirement to be met is that the primary structure be stable and statically determinate.13. For any set of releases there are four new displacement components: the displacement at the thirdrelease resulting from the actual load on the primary structure, and the three flexibility coefficientsf31, f32, and f33. Judicious choice of releases often results in a number of the flexibility coefficients beingzero



Figure 21.12:

Example 21-5: Analysis of Nonprismatic Members, (White et al. 1976)

The nonprismatic beam of Figure 21.13-a is loaded with an end moment MA at its hinged end A.Determine the moment induced at the fixed end B by this loading.

Figure 21.13:

Solution:

1. The beam has one redundant force; we select MB as the redundant R1, and obtain the primarystructure shown in Figure 21.13-c. It can be shown that the flexibility co-efficients for unit momentsapplied at each end are those shown in Fig. 21.13-d and e, with a sign convention of counterclockwiseas positive.2. The equation of consistent displacements at B is

−MAl

8EI+

316

l

EIR1 = 0 (21.43)

and the value of MB is



MB = R1 =23MA (21.44)

3. The resulting moment diagram is given in Figure 21.13-f. We note that the inflection point is 0.40lfrom the fixed end. If the beam had a uniform value of I across its span, the inflection point would beL/3 from the fixed end. Thus the inflection point shifts toward the section of reduced stiffness.4. The end rotation θA is given by

θa =516MAl

EI− 1

8

(23MA

)l

EL=

1148MAl

EI(21.45)

5. The ratio of applied end moment to rotation. MA/θA, is called the rotational stiffness and is

MA

θA=

4811EI

l= 4.364

EI

l(21.46)

6. If we now reverse the boundary conditions, making A fixed and B hinged, and repeat the analysis foran applied moment MB, the resulting moment diagram will be as given in Figure 21.13-h. The momentinduced at end A is only 40% of the applied end moment MB. The inflection point is 0.286l from thefixed end A. The corresponding end rotation θB in Figure 21.13-g is

θB =1180MBl

EI(21.47)

7. The rotational stiffness MB

θBis

MB

θB=

8011EI

l= 7.272

EI

l(21.48)

8. A careful comparison of the rotational stiffnesses, and of the moment diagrams in Figures 21.13-f andh, illustrate the fact that flexural sections of increased stiffness attract more moment, and that inflectionpoints always shift in the direction of decreased stiffness.9. The approach illustrated here may be used to determine moments and end rotations in any type ofnonprismatic member. The end rotations needed in the force analysis may be calculated by either virtualwork or moment area (or by other methods). Complex variations in EI are handled by numerical inte-gration of the virtual work equation or by approximating the resultant M/EI areas and their locationsin the moment area method.

Example 21-6: Fixed End Moments for Nonprismatic Beams, (White et al. 1976)

The beam of example 21-5, with both ends fixed, is loaded with a uniform load w, Figure 21.14-a.Determine the fixed end moments MA and MB.Solution:

1. The beam has two redundant forces and we select MA and MB. Releasing these redundants, R1 andR2, the primary structure is as shown in Figure 21.14-c.2. The equations of consistent deformations are

D1Q + f11R1 + f12R2 = 0 (21.49-a)D2Q + f21R1 + f22R2 = 0 (21.49-b)

where R1 is MA and R2 is MB.3. The values of D1Q and D2Q, the end rotations produced by the real loading on the primary structure,can be computed by the virtual work method.4. The flexibility coefficients are also separately derived (not yet in these notes) and are given in Figures21.14-d and e of the previous example.



Figure 21.14:

5. We define counterclockwise end moments and rotations a positive and obtain

l

EI

[516 − 1

8− 18

316

] [R1

R2

]=wl3

EI

[ −0.352+0.0273

](21.50)

from which

R1 = MA = 0.0742wl2 (21.51-a)

R2 = MB = −0.0961wl2 (21.51-b)

6. The stiffer end of the beam attracts 30% more than the flexible end.7. For a prismatic beam with constant I, the fixed end moments are equal in magnitude (MA = −MB =wl2/12) and intermediate in value between the two end moments determined above.8. Fixed end moments are an essential part of indeterminate analysis based on the displacement (stiff-ness) method and will be used extensively in the Moment Distribution method.

Example 21-7: Rectangular Frame; External Load, (White et al. 1976)

Solution:

1. The structure is statically indeterminate to the third degree, and the displacements (flexibility terms)are shown in Fig. 21.152. In order to evaluate the 9 flexility terms, Fig. 21.16 we refer to Table 21.33. Substituting h = 10 ft, L = 20 ft, and EIb = EIc = EI, the flexibility matrix then becomes

[f ] =1EI

2, 667 3, 000 −300

3, 000 6, 667 −400−300 −400 40

(21.52)

and the vector of displacements for the primary structure is

{D} =1EI

−12, 833−31, 3331, 800

(21.53)

where the units are kips and feet.



Figure 21.15:

Figure 21.16: Definition of Flexibility Terms for a Rigid Frame



DδM

M∫ δ M

Mdx

∫ δ MM EIdx

AB

BC

CD

Tot

al

f 11

✁✁❆❆

✁✁❆❆

+h3 3

+Lh2

+h3 3

+2h3

3EI c

+Lh2

EI b

f 12

✁✁❆❆

✥✥✥

+h2L2

+L2h2

0+

h2L

2EI c

+L2h

2EI b

f 13

✁✁❆❆

−h2 2

−Lh

−h2 2

−h2

EI c

−Lh

EI b

f 21

✥✥✥

✁✁❆❆

+h2L2

+L2h2

0+

h2L

2EI c

+L2h

2EI b

f 22

✥✥✥

✥✥✥

+L2h

+L3 3

0+L2h

EI c

+L3

3EI b

f 23

✥✥✥

−hL

−L2 2

0−

hL

EI c

−L2

2EI b

f 31

✁✁❆❆

−h2 2

−Lh

−h2 2

−h2

EI c

−Lh

EI b

f 32

✥✥✥

−hL

−L2 2

0−

hL

EI c

−L2

2EI b

f 33

+h

+L

+h

+2h

EI c

+LEI b

D1Q

✁✁❆❆

✄✄❛❛❛

−h2(2h+15L−30)

6+Lh(20−5L)

20

−h2(2h+15L−30)

6EI c

+Lh(20−5L)

2EI b

D2Q

✥✥✥

✄✄❛❛❛

−Lh(2h+10L−20)

2+L2(30−10L)

60

−Lh(2h+10L−20)

2EI c

+L2(30−10L)

6EI b

D3Q

✄✄❛❛❛

+h(2h+10L−20)

2−L(20−5L)

20

−h(2h+10L−20)

2EI c

−L(20−5L)

2EI b

Tab

le21

.3:D

ispl

acem

entC

ompu

tation

sfo

ra

Rec

tang

ular

Fram

e



4. The inverse of the flexibility matrix is

[f ]−1 = 10−3EI

2.40 0.00 18.00

0.00 0.375 3.75018.000 3.750 197.5

(21.54)

5. Hence the reactions are determined from

{R} =

R1

R2

R3

= 10−3EI

2.40 0.00 18.00

0.00 0.375 3.75018.000 3.750 197.5

1EI

−12, 833−31, 3331, 800

=

−1.60+5.00−7.00

(21.55)

Example 21-8: Frame with Temperature Effectsand Support Displacements, (White et al. 1976)

The single bay frame, of example 21-7, has a height h = 10 ft and span L = 20 ft and its two suportsrigidly connected and is constructed of reinforced concrete. It supports a roof and wall partitions insuch a manner that a linear temperature variation occurs across the depth of the frame members wheninside and outside temperatures differ. Assume the member depth is constant at 1 ft, and that thestructure was built with fixed bases A and D at a temperature of 85◦F. The temperature is now 70◦Finside and 20◦F outside. We wish to determine the reactions at D under these conditions. Assume thatthe coefficient of linear expansion of reinforced concrete is α = 0.0000055/◦F.Solution:

1. Our analysis proceeds as before, using Equation 21.11 with the [D] vector interpreted appropriately.The three releases shown in Fig. 21.16 will be used.2. The first stage in the analysis is the computation of the relative displacements D1∆, D2∆, D3∆ of theprimary structure caused by temperature effects. These displacements are caused by two effects: axialshortening of the members because of the drop in average temperature (at middepth of the members),and curvature of the members because of the temperature gradient.3. In the following discussion the contributions to displacements due to axial strain are denoted with asingle prime (′) and those due to curvature by a double prime (′′).4. Consider the axial strain first. A unit length of frame member shortens as a result of the temperaturedecrease from 85◦F to 45◦F at the middepth of the member. The strain is therefore

α∆T = (0.0000055)(40) = 0.00022 (21.56)

5. The effect of axial strain on the relative displacements needs little analysis. The horizontal membershortens by an amount (0.00022)(20) = 0.0044 ft. The shortening of the vertical members results in norelative displacement in the vertical direction 2. No rotation occurs.6. We therefore have D′

1∆ = −0.0044 ft, D′2∆ = 0, and D′

3∆ = 0.7. The effect of curvature must also be considered. A frame element of length dx undergoes an angularstrain as a result of the temperature gradient as indicated in Figure 21.17. The change in length at anextreme fiber is

ε = α∆Tdx = 0.0000055(25)dx= 0.000138dx (21.57)

8. with the resulting real rotation of the cross section

dφ = ε/0.5 = 0.000138dx/0.5 = 0.000276dx radians (21.58)

9. The relative displacements of the primary structure at D are found by the virtual force method.10. A virtual force δQ is applied in the direction of the desired displacement and the resulting momentdiagram δM determined.



Figure 21.17:

11. The virtual work equation

δQD =∫δMdφ (21.59)

is used to obtain each of the desired displacements D.12. The results, which you should verify, are

D′′1∆ = 0.0828 ft (21.60-a)

D′′2∆ = 0.1104 ft (21.60-b)

D′′3∆ = −0.01104 radians (21.60-c)

13. Combining the effects of axial and rotational strain, we have

D1∆ = D′1∆ +D′′

1∆ = 0.0784 ft

D2∆ = D′2∆ +D′′

2∆ = 0.1104 ft

D3∆ = D′3∆ +D′′

3∆ = −0.01104 radians

(21.61)

14. We now compute the redundants caused by temperature effects:

[R] = [f ]−1(−[D]) (21.62)

R1

R2

R3

= 10−3EI

[18.0 3.75 197.5

] −0.0784−0.1104+0.01104

=

[+0.0106+0.355

]· 10−3EI (21.63)

where the units are feet and kips.15. You should construct the moment diagram for this structure using the values of the redundantsfound in the analysis.16. Notice that the stiffness term EI does not cancel out in this case. Internal forces and reactionsin a statically indeterminate structure subject to effects other than loads (such as temperature) aredependent on the actual stiffnesses of the structure.17. The effects of axial strain caused by forces in the members have been neglected in this analysis.This is usual for low frames where bending strain dominates behavior. To illustrate the significance ofthis assumption, consider member BC. We have found R1 = 10.6 · 10−6EI k. The tension in BC hasthis same value, resulting in a strain for the member of 10.6 · 10−6EI/EA. For a rectangular member,I/A = (bd3/12)(bd) = d2/12. In our case d = 1 ft, therefore the axial strain is 10.6 · 10−6(0.0833) =8.83 · 10−7, which is several orders of magnitude smaller than the temperature strain computed for thesame member. We may therefore rest assured that neglecting axial strain caused by forces does notaffect the values of the redundants in a significant manner for this structure.18. Now consider the effects of foundation movement on the same structure. The indeterminate framebehavior depends on a structure that we did not design: the earth. The earth is an essential part ofnearly all structures, and we must understand the effects of foundation behavior on structural behavior.



For the purposes of this example, assume that a foundation study has revealed the possibility of aclockwise rotation of the support at D of 0.001 radians and a downward movement of the support at Dof 0.12 ft. We wish to evaluate the redundants R1, R2, and R3 caused by this foundation movement.19. No analysis is needed to determine the values of D1∆, D2∆, and D3∆ for the solution of the redun-dants. These displacements are found directly from the support movements, with proper considerationof the originally chosen sign convention which defined the positive direction of the relative displacements.From the given support displacements, we find D1∆ = 0, D2∆ = +0.12 ft, and D3∆ = −0.001 radians.Can you evaluate these quantities for a case in which the support movements occurred at A instead ofD?20. The values of the redundants is given by

[R] = [f ]−1(−[D]) (21.64) R1

R2

R3

= 10−3EI

[18.0 3.75 197.5

] [ −0.12+0.001

]= 10−6EI

[18.0−252.5

](21.65)

with units in kips and feet.21. A moment diagram may now be constructed, and other internal force quantities computed fromthe now known values of the redundants. The redundants have been valuated separately for effects oftemperature and foundation settlement. These effects may be combined with those due to loading usingthe principle of superposition.

Example 21-9: Braced Bent with Loads and Temperature Change, (White et al. 1976)

The truss shown in Figure 21.18 reperesents an internal braced bent in an enclosed shed, withlateral loads of 20 kN at the panel points. A temperature drop of 30◦C may occur on the outer members(members 1-2, 2-3, 3-4, 4-5, and 5-6). We wish to analyze the truss for the loading and for the temperatureeffect.Solution:

1. The first step in the analysis is the definition of the two redundants. The choice of forces in diagonals2-4 and 1-5 as redundants facilitates the computations because some of the load effects are easy toanalyze. Figure ??-b shows the definition of R1 and R2.2. The computations are organized in tabular form in Table 21.4. The first column gives the bar forcesP in the primary structure caused by the actual loads. Forces are in kN. Column 2 gives the force in eachbar caused by a unit load (1 kN) corresponding to release 1. These are denoted p1 and also represent thebar force q1/δQ1 caused by a virtual force δQ1 applied at the same location. Column 3 lists the samequantity for a unit load and for a virtual force δQ2 applied at release 2. These three columns constitutea record of the truss analysis needed for this problem.3. Column 4 gives the value of L/EA for each bar in terms of Lc/EAc of the vertical members. This isuseful because the term L/EA cancels out in some of the calculations.4. The method of virtual work is applied directly to compute the displacements D1Q and D2Q corre-sponding to the releases and caused by the actual loads. Apply a virtual force δQ1 at release 1. Theinternal virtual forces q1 are found in column 2. The internal virtual work q1∆l is found in column 5 asthe product of columns 1, 2, and 4. The summation of column 5 is D1q = −122.42 Lc/EAc. Similarly,column 6 is the product of columns 1, 3, and 4, giving D2Q = −273.12 Lc/EAc.5. The same method is used to compute the flexibilities fij . In this case the real loading is a unit loadcorresponding first to release 1 leading to f11, and f21, and then to release 2 leading to f12 and f22.Column 7 shows the computation for f11. It is the product of column 2 representing force due to thereal unit load with column 2 representing force due to a virtual load δQ1 at the same location (release1) multiplied by column 4 to include the Lc/EAc term. Column 8 derives from columns 2, 3, and 4 andleads to f21. Columns 9 and 10 are the computations for the remaining flexibilities.



D1Q

D2Q

f11

f21

f12

f22

D1∆

D2∆

Pp1

p2

L/EA

q1PL/EA

q2PL/EA

q1p1L/EA

q2p2L/EA

qp2L/EA

q2p2L/EA

∆/temp

q1∆/

q2∆/

multiply

by

Lc/EAc

Lc/EAc

Lc/EAc

Lc/EAc

Lc/EAc

Lc/EAc

Lc/EAc

Lc

10−4Lc

10−4Lc

1-2

60.0

0−0.707

10

−42.42

00

00.50

−0.0003

02.12

2-3

20.00

−0.707

01

−14.14

00.50

00

0−0.0003

2.12

03-4

0−0.707

02

00

1.00

00

0−0.0003

2.12

04-5

0−0.707

01

00

0.50

00

0−0.0003

2.12

05-6

−20.00

0−0.707

10

14.14

00

00.50

−0.0003

02.12

6-1

40.00

0−0.707

20

−56.56

00

01.00

00

02-5

20.00

−0.707

−0.707

2−28.28

−28.28

1.00

1.00

1.00

1.00

00

01-5

00

1.00

2.828

00

00

02.83

00

02-6

−56.56

01.00

2.828

0−160.00

00

02.83

00

02-4

01.00

02.828

00

2.83

00

00

00

3-5

−28.28

1.00

02.838

−80.00

02.83

00

00

00

−122.42

−273.12

8.66

1.00

1.00

8.66

6.36

4.24



Figure 21.18:

6. We have assumed that a temperature drop of 30◦C occurs in the outer members. The correspondinglength changes are found in column 11. Again using the virtual work method, column 12 tabulates theinternal virtual work of virtual forces q1 through displacements ∆l where for each bar, ∆l = αl∆T .Column 12 is therefore the product of columns 2 and 11. The summation of the elements of column 12is the displacement D1 corresponding to release 1. Column 13 repeats this process for D2 correspondingto release 2.7. The tabulated information provides the necessary terms for a matrix solution of the problem. Wehave

f =[

8.66 1.001.00 8.66

]Lc/EAc (21.66-a)

Dq =[ −122.42−273.12

]Lc/EAc (21.66-b)

D∆ =[

6.364.24

](10)−4Lc (21.66-c)

therefore

f−1 =[

0.117 −0.0134−0.0134 0.117

]EAc/Lc (21.67)

8. The redundant forces due to the applied loading are

R = f−1(−DQ) (21.68-a)

=[ −0.0134 0.117

]EAc/Lc

[122.42273.12

]Lc/EAc =

[10.6630.32

](21.68-b)

9. thus R1 = 10.66 kN, R2 = 30.32 kN.10. The redundant forces due to the temperature drop are R = f−1(−D∆)

=[ −0.0134 0.117

]EAc/Lc

[ −6.36−4.24

]10−4Lc =

[ −6.87−4.11

]10−5EAc

11. Thus with E = 200 kN/mm2, Ac = 500 mm2, we have

R1 = 6.87(10−5)(200)(500) = −6.87kN (21.69-a)R2 = 4.11(10−5)(200)(500) = −4.11kN (21.69-b)



12. Using the redundant forces from each of these analyses, the remainder of the bar forces are computedby simple equilibrium. The information in Table 21.4 contains the basis for such computations. The barforce in any bar is the force of column 1 added to that in column 2 multiplied by R1 plus that in column3 multiplied by R2. This follows from the fact that columns 2 and 3 are bar forces caused by a forceof unity corresponding to each of the redundants. The results of the calculations are shown in Figure??-c for the applied loading and ??-d for the temperature drop. The forces caused by the temperaturedrop are similar in magnitude to those caused by wind load in this example. Temperature differences,shrinkage, support settlement, or tolerance errors can cause important effects in statically indeterminatestructures. These stresses are self-limiting, however, in the sense that if they cause yielding or someductile deformation failure does not necessarily follow, rather relief from th


Draft

Chapter 22

APPROXIMATE FRAMEANALYSIS

1 Despite the widespread availability of computers, approximate methods of analysis are justified by

1. Inherent assumption made regarding the validity of a linear elastic analysis vis a vis of an ultimatefailure design.

2. Ability of structures to redistribute internal forces.

3. Uncertainties in load and material properties

2 Vertical loads are treated separately from the horizontal ones.

3 We use the design sign convention for moments (+ve tension below), and for shear (ccw +ve).

4 Assume girders to be numbered from left to right.

5 In all free body diagrams assume positivee forces/moments, and take algeebraic sums.

6 The key to the approximate analysis method is our ability to sketch the deflected shape of a structureand identify inflection points.

7 We begin by considering a uniformly loaded beam and frame. In each case we consider an extremeend of the restraint: a) free or b) restrained. For the frame a relativly flexible or stiff column would beanalogous to a free or fixed restrain on the beam, Fig. 22.1.

22.1 Vertical Loads

8 With reference to Fig. 22.1, we now consider an intermediary case as shown in Fig. 22.2.

9 With the location of the inflection points identified, we may now determine all the reactions andinternal forces from statics.

10 If we now consider a multi-bay/multi-storey frame, the girders at each floor are assumed to be con-tinuous beams, and columns are assumed to resist the resulting unbalanced moments from the girders,we may make the following assumptions

1. Girders at each floor act as continous beams supporting a uniform load.

2. Inflection points are assumed to be at

(a) One tenth the span from both ends of each girder.

(b) Mid-height of the columns

3. Axial forces and deformation in the girder are negligibly small.

Draft22–2 APPROXIMATE FRAME ANALYSIS

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

0.21 L 0.21 L

LL

+- -

wL/24

w

wL/122

2

Figure 22.1: Uniformly Loaded Beam and Frame with Free or Fixed Beam Restraint



��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

0.5H

0.5H

0.1 L 0.1 L

Figure 22.2: Uniformly Loaded Frame, Approximate Location of Inflection Points



4. Unbalanced end moments from the girders at each joint is distributed to the columns above andbelow the floor.

11 Based on the first assumption, all beams are statically determinate and have a span, Ls equal to 0.8the original length of the girder, L. (Note that for a rigidly connected member, the inflection point isat 0.211 L, and at the support for a simply supported beam; hence, depending on the nature of theconnection one could consider those values as upper and lower bounds for the approximate location ofthe hinge).

12 End forces are given by

Maximum positive moment at the center of each beam is, Fig. 22.3

0.1L

L

0.1L0.8L

Vrgt

Vlft

Mrgt

Mlft

w

Figure 22.3: Approximate Analysis of Frames Subjected to Vertical Loads; Girder Moments

M+ =18wL2s = w

18(0.8)2L2 = 0.08wL2 (22.1)

Maximum negative moment at each end of the girder is given by, Fig. 22.3

M left =M rgt = −w2

(0.1L)2 − w2

(0.8L)(0.1L) = −0.045wL2 (22.2)

Girder Shear are obtained from the free body diagram, Fig. 22.4

V lft =wL

2V rgt = −wL

2(22.3)


Draft22.2 Horizontal Loads 22–5

Vrgti-1 Vlft

i

Pabove

Pbelow

Figure 22.4: Approximate Analysis of Frames Subjected to Vertical Loads; Column Axial Forces

Column axial force is obtained by summing all the girder shears to the axial force transmitted by thecolumn above it. Fig. 22.4

P dwn = Pup + V rgti−1 − V lfti(22.4)

Column Moment are obtained by considering the free body diagram of columns Fig. 22.5

M top =M botabove −M rgt

i−1 +M lfti M bot = −M top (22.5)

Column Shear Points of inflection are at mid-height, with possible exception when the columns onthe first floor are hinged at the base, Fig. 22.5

V =M top

h2

(22.6)

Girder axial forces are assumed to be negligible eventhough the unbalanced column shears above andbelow a floor will be resisted by girders at the floor.

22.2 Horizontal Loads

13 Again, we begin by considering a simple frame subjected to a horizontal force, Fig. 22.6. dependingon the boundary conditions, we will have different locations for the inflection points.

14 For the analysis of a multi-bays/multi-storeys frame, we must differentiate between low and high risebuildings.

Low rise buidlings, where the height is at least samller than the hrizontal dimension, the deflectedshape is characterized by shear deformations.



h/2

h/2

h/2

h/2

Mcolbelow

Mi-1rgt Mi

lft

Mcolabove

Li

h/2

h/2

h/2

h/2

Li-1

Vi-1lft Vi

lft Virgt

Vi-1rgt

Mi-1lft

Mirgt

Figure 22.5: Approximate Analysis of Frames Subjected to Vertical Loads; Column Moments



PHL

PHL

2P

2P

��

��

L

HI

Iα2 2

PH PH

2P

2P

��

��

��

��

PH2L

PH2L

PH4

PH4

PH4

PH4

PH4

��

��

L

HI

Iα

PHP

P

4

Figure 22.6: Horizontal Force Acting on a Frame, Approximate Location of Inflection Points



High rise buildings, where the height is several times greater than its least horizontal dimension, thedeflected shape is dominated by overall flexural deformation.

22.2.1 Portal Method

15 Low rise buildings under lateral loads, have predominantly shear deformations. Thus, the approximateanalysis of this type of structure is based on

1. Distribution of horizontal shear forces.

2. Location of inflection points.

16 The portal method is based on the following assumptions

1. Inflection points are located at

(a) Mid-height of all columns above the second floor.

(b) Mid-height of floor columns if rigid support, or at the base if hinged.

(c) At the center of each girder.

2. Total horizontal shear at the mid-height of all columns at any floor level will be distributed amongthese columns so that each of the two exterior columns carry half as much horizontal shear as eachinterior columns of the frame.

17 Forces are obtained from

Column Shear is obtained by passing a horizontal section through the mid-height of the columns ateach floor and summing the lateral forces above it, then Fig. 22.7

H/2 H/2HH

Figure 22.7: Approximate Analysis of Frames Subjected to Lateral Loads; Column Shear

V ext =∑F lateral

2No. of baysV int = 2V ext (22.7)



Column Moments at the end of each column is equal to the shear at the column times half the heightof the corresponding column, Fig. 22.7

M top = Vh

2M bot = −M top (22.8)

Girder Moments is obtained from the columns connected to the girder, Fig. 22.8

h/2

h/2

h/2

h/2

Mcolbelow

Mi-1rgt Mi

lft

Mcolabove

Li-1/2 Li/2h/2

h/2

h/2

h/2

Li/2Li-1/2

Vi-1lft Vi

lft Virgt

Vi-1rgt

Mi-1lft

Mirgt

Figure 22.8: ***Approximate Analysis of Frames Subjected to Lateral Loads; Girder Moment

M lfti =Mabove

col −M belowcol +M rgt

i−1 M rgti = −M lft

i(22.9)

Girder Shears Since there is an inflection point at the center of the girder, the girder shear is obtainedby considering the sum of moments about that point, Fig. 22.8

V lft = −2ML

V rgt = V lft (22.10)

Column Axial Forces are obtained by summing girder shears and the axial force from the columnabove, Fig. ??

P = P above + P rgt + P lft (22.11)



Vrgti-1 Vlft

i

Pabove

Pbelow

Figure 22.9: Approximate Analysis of Frames Subjected to Lateral Loads; Column Axial Force

18 In either case, you should always use a free body diagram in conjunction with this method, andnever rely on a blind application of the formulae.

Example 22-1: Approximate Analysis of a Frame subjected to Vertical and Horizontal Loads

Draw the shear, and moment diagram for the following frame. Solution:

15K

30K

14’14’

16’16’

20’20’ 30’30’ 24’24’

5 6

1

12 13 14

11109

87

3 42

0.25K/ft

0.50K/ft

Figure 22.10: Example; Approximate Analysis of a Building

Vertical LoadsThe analysis should be conducted in conjunction with the free body diagram shown in Fig. 22.11.



0.56 0.7 0.45 0.81

4.5 5.6 3.6 6.5

4.5 5.6 3.6 6.5

0.64 0.8 0.930.51

4.5 5.6 3.6 3.6

5.64.5 3.6 6.5

9.0

20.2

20.2

13.09.

0

13.0

0.64

4.5

10.1

6.5

10.14.

5

6.5

7.5

7.5 6.0

6.0

2.5

2.5

3.75

3.75

0.8 0.51 0.93

5.0

5.0

3.0

3.0

0.56 0.7 0.45 0.81

Figure 22.11: Free Body Diagram for the Approximate Analysis of a Frame Subjected to Vertical Loads



1. Top Girder Moments

M lft12 = −0.045w12L212 = −(0.045)(0.25)(20)2 = − 4.5 k.ft

M cnt12 = 0.08w12L212 = (0.08)(0.25)(20)2 = 8.0 k.ft

M rgt12 = M lft

12 = − 4.5 k.ft

M lft13 = −0.045w13L213 = −(0.045)(0.25)(30)2 = − 10.1 k.ft

M cnt13 = 0.08w13L213 = (0.08)(0.25)(30)2 = 18.0 k.ft

M rgt13 = M lft

13 = − 10.1 k.ft

M lft14 = −0.045w14L214 = −(0.045)(0.25)(24)2 = − 6.5 k.ft

M cnt14 = 0.08w14L214 = (0.08)(0.25)(24)2 = 11.5 k.ft

M rgt14 = M lft

14 = − 6.5 k.ft

2. Bottom Girder Moments

M lft9 = −0.045w9L29 = −(0.045)(0.5)(20)2 = − 9.0 k.ft

M cnt9 = 0.08w9L29 = (0.08)(0.5)(20)2 = 16.0 k.ft

M rgt9 = M lft

9 = − 9.0 k.ft

M lft10 = −0.045w10L210 = −(0.045)(0.5)(30)2 = − 20.3 k.ft

M cnt10 = 0.08w10L210 = (0.08)(0.5)(30)2 = 36.0 k.ft

M rgt10 = M lft

11 = − 20.3 k.ft

M lft11 = −0.045w12L212 = −(0.045)(0.5)(24)2 = − 13.0 k.ft

M cnt11 = 0.08w12L212 = (0.08)(0.5)(24)2 = 23.0 k.ft

M rgt11 = M lft

12 = − 13.0 k.ft

3. Top Column Moments

M top5 = +M lft

12 = − 4.5 k.ft

M bot5 = −M top

5 = 4.5 k.ft

M top6 = −M rgt

12 +M lft13 = −(−4.5) + (−10.1) = − 5.6 k.ft

M bot6 = −M top

6 = 5.6 k.ft

M top7 = −M rgt

13 +M lft14 = −(−10.1) + (−6.5) = − 3.6 k.ft

M bot7 = −M top

7 = 3.6 k.ft

M top8 = −M rgt

14 = −(−6.5) = 6.5 k.ft

M bot8 = −M top

8 = − 6.5 k.ft

4. Bottom Column Moments

M top1 = +M bot

5 +M lft9 = 4.5− 9.0 = − 4.5 k.ft

M bot1 = −M top

1 = 4.5 k.ft

M top2 = +M bot

6 −M rgt9 +M lft

10 = 5.6− (−9.0) + (−20.3) = − 5.6 k.ft

M bot2 = −M top

2 = 5.6 k.ft

M top3 = +M bot

7 −M rgt10 +M lft

11 = −3.6− (−20.3) + (−13.0) = 3.6 k.ft

M bot3 = −M top

3 = − 3.6 k.ft

M top4 = +M bot

8 −M rgt11 = −6.5− (−13.0) = 6.5 k.ft

M bot4 = −M top

4 = − 6.5 k.ft

5. Top Girder ShearV lft12 = w12L12

2 = (0.25)(20)2 = 2.5 k

V rgt12 = −V lft12 = − 2.5 k

V lft13 = w13L132 = (0.25)(30)

2 = 3.75 k

V rgt13 = −V lft13 = − 3.75 k

V lft14 = w14L142 = (0.25)(24)

2 = 3.0 k

V rgt14 = −V lft14 = − 3.0 k



0.25K/ft

0.50K/ft 14’14’

16’16’

20’20’ 30’30’ 24’24’

5 6

1

12 13 14

11109

87

3 42

-9.0’-9.0’k

-4.5’-4.5’k

-10.1’-10.1’k -10.1’-10.1’k

-13.0’-13.0’k

-4.5’-4.5’k

-4.5’-4.5’k

+4.5’+4.5’k

+4.5’+4.5’k +5.6’+5.6’k

+5.6’+5.6’k

-5.6’-5.6’k

-5.6’-5.6’k +3.6’+3.6’k

+3.6’+3.6’k

-3.6’-3.6’k

-3.6’-3.6’k

-6.5’-6.5’k

-6.5’-6.5’k

+6.5’+6.5’k

+6.5’+6.5’k

-13.0’-13.0’k

-20.2’-20.2’k-20.2’-20.2’k

+8.0’+8.0’k +18.0’+18.0’k

+11.5’+11.5’k

+16.0’+16.0’k

+32.0’+32.0’k+23.0’+23.0’k

-6.5’-6.5’k-6.5’-6.5’k

-4.5’-4.5’k

-9.0’-9.0’k

Figure 22.12: Approximate Analysis of a Building; Moments Due to Vertical Loads



6. Bottom Girder Shear

V lft9 = w9L92 = (0.5)(20)

2 = 5.00 k

V rgt9 = −V lft9 = − 5.00 k

V lft10 = w10L102 = (0.5)(30)

2 = 7.50 k

V rgt10 = −V lft10 = − 7.50 k

V lft11 = w11L112 = (0.5)(24)

2 = 6.00 k

V rgt11 = −V lft11 = − 6.00 k

7. Column ShearsV5 = Mtop

5H52

= −4.5142

= − 0.64 k

V6 = Mtop6H62

= −5.6142

= − 0.80 k

V7 = Mtop7H72

= 3.6142

= 0.52 k

V8 = Mtop8H82

= 6.5142

= 0.93 k

V1 = Mtop1H12

= −4.5162

= − 0.56 k

V2 = Mtop2H22

= −5.6162

= − 0.70 k

V3 = Mtop3H32

= 3.6162

= 0.46 k

V4 = Mtop4H42

= 6.5162

= 0.81 k

8. Top Column Axial Forces

P5 = V lft12 = 2.50 k

P6 = −V rgt12 + V lft13 = −(−2.50) + 3.75 = 6.25 k

P7 = −V rgt13 + V lft14 = −(−3.75) + 3.00 = 6.75 k

P8 = −V rgt14 = 3.00 k

9. Bottom Column Axial Forces

P1 = P5 + V lft9 = 2.50 + 5.0 = 7.5 k

P2 = P6 − V rgt10 + V lft9 = 6.25− (−5.00) + 7.50 = 18.75 k

P3 = P7 − V rgt11 + V lft10 = 6.75− (−7.50) + 6.0 = 20.25 k

P4 = P8 − V rgt11 = 3.00− (−6.00) = 9.00 k

Horizontal Loads, Portal Method

1. Column ShearsV5 = 15

(2)(3) = 2.5 k

V6 = 2(V5) = (2)(2.5) = 5 k

V7 = 2(V5) = (2)(2.5) = 5 k

V8 = V5 = 2.5 k

V1 = 15+30(2)(3) = 7.5 k

V2 = 2(V1) = (2)(7.5) = 15 k

V3 = 2(V1) = (2)(2.5) = 15 k

V4 = V1 = 7.5 k



+2.5K +3.75K

-3.75K

+3.0K

-3.0K

+5.0K

-5.0K

+7.5K

-7.5K

-0.64K -0.80K +0.51K +0.93K

+0.81K+0.45K-0.70K-0.56K

-6.0K

+6.0K

-2.5K

Figure 22.13: Approximate Analysis of a Building; Shears Due to Vertical Loads



Approximate Analysis Vertical Loads APROXVER.XLS Victor E. Saouma

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123456789

101112131415161718192021222324252627282930

A B C D E F G H I J K L M N O P Q

L1 L2 L3Height Span 20 30 24

14 Load 0.25 0.25 0.2516 Load 0.5 0.5 0.5

MOMENTSBay 1 Bay 2 Bay 3

Col Beam Column Beam Column Beam ColLft Cnt Rgt Lft Cnr Rgt Lft Cnt Rgt

-4.5 8.0 -4.5 -10.1 18.0 -10.1 -6.5 11.5 -6.5-4.5 -5.6 3.6 6.54.5 5.6 -3.6 -6.5

-9.0 16.0 -9.0 -20.3 36.0 -20.3 -13.0 23.0 -13.0-4.5 -5.6 3.6 6.54.5 5.6 -3.6 -6.5

SHEARBay 1 Bay 2 Bay 3

Col Beam Column Beam Column Beam ColLft Rgt Lft Rgt Lft Rgt

2.50 -2.50 3.75 -3.75 3.00 -3.00-0.64 -0.80 0.52 0.93

5.00 -5.00 7.50 -7.50 6.00 -6.00-0.56 -0.70 0.46 0.81

AXIAL FORCEBay 1 Bay 2 Bay 3

Col Beam Column Beam Column Beam Col0.00 0.00 0.00

2.50 6.25 6.75 3.000.00 0.00 0.00

7.50 18.75 20.25 9.00

Figure 22.14: Approximate Analysis for Vertical Loads; Spread-Sheet Format



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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

AB

CD

EF

GH

IJ

KL

MN

OP

Q

L1L2

L3H

eig

htSp

an

2030

2414

Loa

d0.

250.

250.

2516

Loa

d0.

50.

50.

5M

OM

ENTS

Bay

1Ba

y 2

Bay

3C

ol

Bea

mC

olu

mn

Bea

mC

olu

mn

Bea

mC

ol

Lft

Cn

tR

gt

Lft

Cn

rR

gt

Lft

Cn

tR

gt

=-0

.045

*D4*

D3^

2=

0.08

*D4*

D3*

D3

=+

D10

=-0

.045

*I4*

I3^

2=

0.08

*I4*

I3*I

3=

+I1

0=

-0.0

45*N

4*N

3^2

=0.

08*N

4*N

3*N

3=

N10

=+

D10

=-F

10+

I10

=-K

10+

N10

=-P

10=

-C11

=-G

11=

-L11

=-Q

11

=-0

.045

*D5*

D3^

2=

0.08

*D5*

D3*

D3

=+

D13

=-0

.045

*I5*

I3^

2=

0.08

*I5*

I3*I

3=

+I1

3=

-0.0

45*N

5*N

3^2

=0.

08*N

5*N

3*N

3=

+N

13

=+

D13

+C

12=

-F13

+I1

3+G

12=

-K13

+N

13+

L12

=-P

13+

Q12

=-C

14=

-G14

=-L

14=

-Q14

SHEA

RBa

y 1

Bay

2Ba

y 3

Co

lBe

am

Co

lum

nBe

am

Co

lum

nBe

am

Co

l

Lft

Rg

tLf

tR

gt

Lft

Rg

t

=+

D3*

D4/

2=

-D20

=+

I3*I

4/2

=-I2

0=

+N

3*N

4/2

=-N

20

=2*

C11

/A4

=2*

G11

/A4

=2*

L11/

A4

=2*

Q11

/A4

=+

D3*

D5/

2=

-D22

=+

I3*I

5/2

=-I2

2=

+N

3*N

5/2

=-N

22

=2*

C14

/A5

=2*

G14

/A5

=2*

L14/

A5

=2*

Q14

/A5

AX

IAL

FO

RCE

Bay

1Ba

y 2

Bay

3C

ol

Bea

mC

olu

mn

Bea

mC

olu

mn

Bea

mC

ol

00

0

=+

D20

=-F

20+

I20

=-K

20+

N20

=-P

20

00

0

=+

C28

+D

22=

+G

28-F

22+

I22

=+

L28-

K22+

N22

=+

Q28

-P22

Figure 22.15: Approximate Analysis for Vertical Loads; Equations in Spread-Sheet



17.5

17.5

17.5

17.5

17.5

17.5

17.5

77.5

77.5

77.5

77.5

77.5

77.5

35 35 17.5

17.5 35 35 17.5

60 120 120 60

60 120120 60

15

30

2.5 5 5 2.5

7.5 15 15 7.5

7.515157.5

2.5 5 5 2.5

Figure 22.16: Free Body Diagram for the Approximate Analysis of a Frame Subjected to Lateral Loads



2. Top Column Moments

M top5 = V1H5

2 = (2.5)(14)2 = 17.5 k.ft

M bot5 = −M top

5 = − 17.5 k.ft

M top6 = V6H6

2 = (5)(14)2 = 35.0 k.ft

M bot6 = −M top

6 = − 35.0 k.ft

M top7 = V up7 H7

2 = (5)(14)2 = 35.0 k.ft

M bot7 = −M top

7 = − 35.0 k.ft

M top8 = V up8 H8

2 = (2.5)(14)2 = 17.5 k.ft

M bot8 = −M top

8 = − 17.5 k.ft

3. Bottom Column Moments

M top1 = V dwn1 H1

2 = (7.5)(16)2 = 60 k.ft

M bot1 = −M top

1 = − 60 k.ft

M top2 = V dwn2 H2

2 = (15)(16)2 = 120 k.ft

M bot2 = −M top

2 = − 120 k.ft

M top3 = V dwn3 H3

2 = (15)(16)2 = 120 k.ft

M bot3 = −M top

3 = − 120 k.ft

M top4 = V dwn4 H4

2 = (7.5)(16)2 = 60 k.ft

M bot4 = −M top

4 = − 60 k.ft

4. Top Girder Moments

M lft12 = M top

5 = 17.5 k.ft

M rgt12 = −M lft

12 = − 17.5 k.ft

M lft13 = M rgt

12 +M top6 = −17.5 + 35 = 17.5 k.ft

M rgt13 = −M lft

13 = − 17.5 k.ft

M lft14 = M rgt

13 +M top7 = −17.5 + 35 = 17.5 k.ft

M rgt14 = −M lft

14 = − 17.5 k.ft

5. Bottom Girder Moments

M lft9 = M top

1 −M bot5 = 60− (−17.5) = 77.5 k.ft

M rgt9 = −M lft

9 = − 77.5 k.ft

M lft10 = M rgt

9 +M top2 −M bot

6 = −77.5 + 120− (−35) = 77.5 k.ft

M rgt10 = −M lft

10 = − 77.5 k.ft

M lft11 = M rgt

10 +M top3 −M bot

7 = −77.5 + 120− (−35) = 77.5 k.ft

M rgt11 = −M lft

11 = − 77.5 k.ft

6. Top Girder Shear

V lft12 = − 2Mlft12

L12= − (2)(17.5)

20 = −1.75 k

V rgt12 = +V lft12 = −1.75 k


L13= − (2)(17.5)

30 = −1.17 k

V rgt13 = +V lft13 = −1.17 k


L14= − (2)(17.5)

24 = −1.46 k

V rgt14 = +V lft14 = −1.46 k

7. Bottom Girder Shear


L9= − (2)(77.5)

20 = −7.75 k

V rgt9 = +V lft9 = −7.75 k


L10= − (2)(77.5)

30 = −5.17 k

V rgt10 = +V lft10 = −5.17 k


L11= − (2)(77.5)

24 = −6.46 k

V rgt11 = +V lft11 = −6.46 k



+17.5’K+17.5’K

+17.5’K

-17.5’K

+60’K

+60’K+120’K

-120’K -120’K

+120’K

-60’K

-60’K

-17.5’K

+17.5’K+35’K

-35’K -35’K

+35’K

+17.5’K

+77.5’K +77.5’K +77.5’K

-77.5’K -77.5’K -77.5’K

-17.5K-17.5K -17.5K

14’

16’

20’ 30’ 24’

5 6

1

12 13 14

11109

87

3 42

15K

30K

Figure 22.17: Approximate Analysis of a Building; Moments Due to Lateral Loads



8. Top Column Axial Forces (+ve tension, -ve compression)

P5 = −V lft12 = −(−1.75) k

P6 = +V rgt12 − V lft13 = −1.75− (−1.17) = −0.58 k

P7 = +V rgt13 − V lft14 = −1.17− (−1.46) = 0.29 k

P8 = V rgt14 = −1.46 k

9. Bottom Column Axial Forces (+ve tension, -ve compression)

P1 = P5 + V lft9 = 1.75− (−7.75) = 9.5 k

P2 = P6 + V rgt10 + V lft9 = −0.58− 7.75− (−5.17) = −3.16 k

P3 = P7 + V rgt11 + V lft10 = 0.29− 5.17− (−6.46) = 1.58 k

P4 = P8 + V rgt11 = −1.46− 6.46 = −7.66 k

Design Parameters On the basis of the two approximate analyses, vertical and lateral load, we nowseek the design parameters for the frame, Table 22.2.

Mem. Vert. Hor. DesignValues

Moment 4.50 60.00 64.501 Axial 7.50 9.50 17.00

Shear 0.56 7.50 8.06Moment 5.60 120.00 125.60

2 Axial 18.75 15.83 34.58Shear 0.70 15.00 15.70Moment 3.60 120.00 123.60






8 Axial 3.00 1.46 4.46Shear 0.93 2.50 3.43

Table 22.1: Columns Combined Approximate Vertical and Horizontal Loads



Portal Method PORTAL.XLS Victor E. Saouma





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123456789

101112131415161718192021222324252627282930

A B C D E F G H I J K L M N O P Q R SPORTAL METHOD

# of Bays 3 L1 L2 L320 30 24

MOMENTS# of Storeys 2 Bay 1 Bay 2 Bay 3

Force Shear Col Beam Column Beam Column Beam ColH Lat. Tot Ext Int Lft Rgt Lft Rgt Lft Rgt

17.5 -17.5 17.5 -17.5 17.5 -17.5H1 14 15 15 2.5 5 17.5 35.0 35.0 17.5

-17.5 -35.0 -35.0 -17.577.5 -77.5 77.5 -77.5 77.5 -77.5

H2 16 30 45 7.5 15 60.0 120.0 120.0 60.0-60.0 -120.0 -120.0 -60.0

SHEARBay 1 Bay 2 Bay 3

Col Beam Column Beam Column Beam ColLft Rgt Lft Rgt Lft Rgt

-1.75 -1.75 -1.17 -1.17 -1.46 -1.462.50 5.00 5.00 2.502.50 5.00 5.00 2.50

-7.75 -7.75 -5.17 -5.17 -6.46 -6.467.50 15.00 15.00 7.507.50 15.00 15.00 7.50

AXIAL FORCEBay 1 Bay 2 Bay 3

Col Beam Column Beam Column Beam Col0.00 0.00 0.00

1.75 -0.58 0.29 -1.460.00 0.00 0.00

9.50 -3.17 1.58 -7.92

Figure 22.18: Portal Method; Spread-Sheet Format



Portal Method PORTAL.XLS Victor E. Saouma

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1

23

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67

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910

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1213

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1920

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2223

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A B C D E F G H I J K L M N O P Q R S

PORTAL METHOD# of Bays 3 L1 L2 L3

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20 30 24

MOMENTSAAAAAAAA

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# of Storeys 2 Bay 1 Bay 2 Bay 3AAAA

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Force Shear Col Beam Column Beam Column Beam ColAAAA

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H Lat. Tot Ext Int Lft Rgt Lft Rgt Lft Rgt

=+H9 =-I8 =+J8+K9 =-M8 =+N8+O9 =-Q8AAAAAAAA

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H1 14 15 =+C9 =+D9/(2*$F$2) =2*E9 =+E9*B9/2 =+F9*B9/2 =+K9 =+H9

=-H9 =-K9 =+K10 =+H10

=+H12-H10 =-I11 =+K12-K10+J11 =-M11 =+O12-O10+N11 =-Q11AAAAAAAAAAA

AAAAAAAAAAA

H2 16 30 =SUM($C$9:C12) =+D12/(2*$F$2) =2*E12 =+E12*B12/2 =+F12*B12/2 =+K12 =+H12

=-H12 =-K12 =+K13 =+H13

SHEARAAAAAAAAAAAA

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Col Beam Column Beam Column Beam ColAAAA

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Lft Rgt Lft Rgt Lft Rgt

=-2*I8/I$3 =+I18 =-2*M8/M$3 =+M18 =-2*Q8/Q$3 =+Q18AAAA

=+E9 =+F9 =+F9 =+E9

=+H19 =+K19 =+O19 =+S19

=-2*I11/I$3 =+I21 =-2*M11/M$3 =+M21 =-2*Q11/Q$3 =+Q21

=+E12 =+F12 =+F12 =+E12

=+H22 =+K22 =+O22 =+S22

AXIAL FORCE

Bay 1 Bay 2 Bay 3AAAAA

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Col Beam Column Beam Column Beam ColAAAA

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0 0 0

=-I18 =+J18-M18 =+N18-Q18 =+R18

0 0 0

=+H28-I21 =+K28+J21-M21 =+O28+N21-Q21 =+S28+R21AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA





Figure 22.19: Portal Method; Equations in Spread-Sheet



Mem. Vert. Hor. DesignValues

-ve Moment 9.00 77.50 86.509 +ve Moment 16.00 0.00 16.00

Shear 5.00 7.75 12.75-ve Moment 20.20 77.50 97.70

10 +ve Moment 36.00 0.00 36.00Shear 7.50 5.17 12.67-ve Moment 13.0 77.50 90.50




14 +ve Moment 11.50 0.00 11.50Shear 3.00 1.46 4.46

Table 22.2: Girders Combined Approximate Vertical and Horizontal Loads


Draft

Chapter 23

KINEMATICINDETERMINANCY; STIFFNESSMETHOD

23.1 Introduction

23.1.1 Stiffness vs Flexibility

1 There are two classes of structural analysis methods, Table 23.1:

Flexibility: where the primary unknown is a force, where equations of equilibrium are the startingpoint, static indeterminancy occurs if there are more unknowns than equations, and displacementsof the entire structure (usually from virtual work) are used to write an equation of compatibilityof displacements in order to solve for the redundant forces.

Stiffness: method is the counterpart of the flexibility one. Primary unknowns are displacements, andwe start from expressions for the forces written in terms of the displacements (at the elementlevel) and then apply the equations of equilibrium. The structure is considered to be kinematicallyindeterminate to the nth degree where n is the total number of independent displacements. Fromthe displacements, we then compute the internal forces.

Flexibility StiffnessPrimary Variable (d.o.f.) Forces DisplacementsIndeterminancy Static KinematicForce-Displacement Displacement(Force)/Structure Force(Displacement)/ElementGoverning Relations Compatibility of displacement EquilibriumMethods of analysis “Consistent Deformation” Slope Deflection; Moment Distribution

Table 23.1: Stiffness vs Flexibility Methods

2 In the flexibility method, we started by releasing as many redundant forces as possible in order to renderthe structure statically determinate, and this made it quite flexible. We then applied an appropriate setof forces such that kinematic constraints were satisifed.

3 In the stiffness method, we follow a different approach, we stiffen the structure by constraining all thedisplacements, hence making it kinematically determinate, and then we will release all the constraintsin such a way to satisfy equilibrium.

Draft23–2 KINEMATIC INDETERMINANCY; STIFFNESS METHOD

4 In the slope deflection method, all constraints are released simultaneously, thus resulting in a linearsystem of n equations with n unknowns. In the Moment Distribution method, we release the constraintsone at a time and essentially solve for the system of n equations iteratively.

23.1.2 Sign Convention

5 The sign convention in the stiffness method is different than the one previously adopted in structuralanalysis/design, Fig. 26.3.

6 In the stiffness method the sign convention adopted is consistent with the prevailing coordinate system.Hence, we define a positive moment as one which is counter-clockwise at the end of the element, Fig.26.3.

Figure 23.1: Sign Convention, Design and Analysis

23.2 Degrees of Freedom

7 A degree of freedom (d.o.f.) is an independent generalized nodal displacement of a node.

8 The displacements must be linearly independent and thus not related to each other. For example, aroller support on an inclined plane would have three displacements (rotation θ, and two translations uand v), however since the two displacements are kinematically constrained, we only have two independentdisplacements, Fig. 26.5.

Figure 23.2: Independent Displacements

9 We note that we have been referring to generalized displacements, because we want this term to includetranslations as well as rotations. Depending on the type of structure, there may be none, one or morethan one such displacement.

10 The types of degrees of freedom for various types of structures are shown in Table 26.4


Draft23.3 Kinematic Relations 23–3

Type Node 1 Node 2

1 Dimensional{p} Fy1, Mz2 Fy3, Mz4

Beam{δ} v1, θ2 v3, θ4

2 Dimensional{p} Fx1 Fx2

Truss{δ} u1 u2{p} Fx1, Fy2, Mz3 Fx4, Fy5, Mz6

Frame{δ} u1, v2, θ3 u4, v5, θ6

3 Dimensional{p} Fx1, Fx2

Truss{δ} u1, u2{p} Fx1, Fy2, Fy3, Fx7, Fy8, Fy9,

Tx4 My5, Mz6 Tx10 My11, Mz12

Frame{δ} u1, v2, w3, u7, v8, w9,

θ4, θ5 θ6 θ10, θ11 θ12

Table 23.2: Degrees of Freedom of Different Structure Types Systems

11 Fig. 26.4 also shows the geometric (upper left) and elastic material (upper right) properties associatedwith each type of element.

23.2.1 Methods of Analysis

12 There are three methods for the stiffness based analysis of a structure

Slope Deflection: (Mohr, 1892) Which results in a system of n linear equations with n unknowns,where n is the degree of kinematic indeterminancy (i.e. total number of independent displace-ments/rotation).

Moment Distribution: (Cross, 1930) which is an iterative method to solve for the n displacementsand corresponding internal forces in flexural structures.

Direct Stiffness method: (1960) which is a formal statement of the stiffness method and cast inmatrix form is by far the most powerful method of structural analysis.

The first two methods lend themselves to hand calculation, and the third to a computer based analysis.

23.3 Kinematic Relations

23.3.1 Force-Displacement Relations

13 Whereas in the flexibility method we sought to obtain a displacement in terms of the forces (throughvirtual work) for an entire structure, our starting point in the stiffness method is to develop a set ofrelationship for the force in terms of the displacements for a single element.

V1M1

V2M2

=

− − − −− − − −− − − −− − − −

v1θ1v2θ2

(23.1)



Figure 23.3: Total Degrees of Freedom for various Type of Elements



14 We start from the differential equation of a beam, Fig. 23.4 in which we have all positive knowndisplacements, we have from strength of materials

θ

θ

V V

M

M

vv

L

1

1

1

1 2

2

2

2

1 2

Figure 23.4: Flexural Problem Formulation

M = −EI d2v

dx2=M1 − V1x+m(x) (23.2)

where m(x) is the moment applied due to the applied load only. It is positive when counterclockwise.

15 Integrating twice

−EIv′ = M1x− 12V1x

2 + f(x) + C1 (23.3)

−EIv =12M1x

2 − 16V1x

3 + g(x) + C1x+ C2 (23.4)

where f(x) =∫m(x)dx, and g(x) =

∫f(x)dx.

16 Applying the boundary conditions at x = 0

v′ = θ1v = v1

}⇒{C1 = −EIθ1C2 = −EIv1 (23.5)

17 Applying the boundary conditions at x = L and combining with the expressions for C1 and C2

v′ = θ2v = v2

}⇒{ −EIθ2 = M1L− 1

2V1L2 + f(L)− EIθ1

−EIv2 = 12M1L

2 − 16V1L

3 + g(L)− EIθ1L− EIv1 (23.6)

18 Since equilibrium of forces and moments must be satisfied, we have:

V1 + q + V2 = 0 M1 − V1L+m(L) +M2 = 0 (23.7)

where q =∫ L0p(x)dx, thus

V1 =(M1 +M2)

L+

1Lm(L) V2 = −(V1 + q) (23.8)

19 Substituting V1 into the expressions for θ2 and v2 in Eq. 23.6 and rearranging{M1 −M2 = 2EIz

L θ1 + 2EIzL θ2 +m(L)− 2

Lf(L)2M1 −M2 = 6EIz

L θ1 − 6EIzL2 v1 − 6EIz

L2 v2 +m(L)− 6L2 g(L)

(23.9)

20 Solving those two equations, we obtain:



M1 =2EIzL

(2θ1 + θ2)− 6EIzL2

(v2 − v1)︸︷︷︸I

+MF1︸︷︷︸

II

(23.10)

M2 =2EIzL

(θ1 + 2θ2)− 6EIzL2

(v2 − v1)︸︷︷︸I

+MF2︸︷︷︸

II

(23.11)

where

MF1 =

2L2

[Lf(L)− 3g(L)] (23.12-a)

MF2 = − 1

L2[L2m(L)− 4Lf(L) + 6g(L)

](23.12-b)

MF1 and MF

2 are the fixed end moments for θ1 = θ2 = 0 and v1 = v2 = 0, that is fixed end moments.They can be obtained either from the analysis of a fixed end beam, or more readily from the precedingtwo equations.

21 In Eq. 23.10 and 23.11 we observe that the moments developed at the end of a member are causedby: I) end rotation and displacements; and II) fixed end members.

22 Finally, we can substitute those expressions in Eq. 23.8

V1 =6EIzL2

(θ1 + θ2)− 12EIzL3

(v2 − v1)︸︷︷︸I

+ V F1︸︷︷︸II

(23.13)

V2 = −6EIzL2

(θ1 + θ2) +12EIzL3

(v2 − v1)︸︷︷︸I

+ V F2︸︷︷︸II

(23.14)

where

V F1 =6L3

[Lf(L)− 2g(L)] (23.15-a)

V F2 = −[

6L3

[Lf(L)− 2g(L)] + q]

(23.15-b)

23 The relationships just derived enable us now to determine the stiffness matrix of a beam element.

V1M1

V2M2

=

v1 θ1 v2 θ2V1

12EIzL3

6EIzL2 − 12EIz

L36EIzL2

M16EIzL2

4EIzL − 6EIz

L22EIzL

V2 − 12EIzL3 − 6EIz

L212EIzL3 − 6EIz

L2

M26EIzL2

2EIzL − 6EIz

L24EIzL

︸︷︷︸ke

v1θ1v2θ2

(23.16)

23.3.2 Fixed End Actions

24 As mentioned above, the end actions developed in a member involve the end displacements, rotations,and the in-span loads. In-spans loads exhibit themselves in the form of fixed-end forces.

25 The fixed-end actions can be determined from the equations derived above, or by analyzing a fixed-endbeam under the applied loads.

26 Note that in both cases, the load has to be assumed positive, i.e. pointing up for a beam.

27 The equations derived for calculating the fixed-end actions can be summarized as follows. We recallthat, with the x axis directed to the right, positive loads and shear forces act upward and positive



moments are counterclockwise. To calculate the fixed end actions the only thing we need is an expressionfor the moment of the applied loads (without the end reactions) in the analysis sign convention. Thuswith

m(x) = moment due to the applied loads at section x (23.17-a)

f(x) =∫m(x)dx (23.17-b)

g(x) =∫f(x)dx (23.17-c)

q =∫p(x)dx = total load on the span (23.17-d)

and

MF1 =

2L2

[Lf(L)− 3g(L)] (23.18)

MF2 = − 1

L2[L2m(L)− 4Lf(L) + 6g(L)

](23.19)

V F1 =6L3

[Lf(L)− 2g(L)] (23.20)

V F2 = − 6L3

[Lf(L)− 2g(L)]− q (23.21)

23.3.2.1 Uniformly Distributed Loads

28 For a uniformly distributed load w over the entire span,

m(x) = −12wx2; f(x) = −1

6wx3; g(x) = − 1

24wx4; q = wL (23.22)

29 Substituting

MF1 =

2L2

[L

(−1

6wL3

)− 3

(− 1

24wL4

)]= −wL2

12 (23.23-a)

MF2 = − 1

L2

[L2(−1

2wL2

)− 4L

(−1

6wL3

)+ 6

(− 1

24wL4

)]= wL2

12 (23.23-b)

V F1 =6L3

[L

(−1

6wL3

)− 2

(− 1

24wL4

)]= −wL

2 (23.23-c)

V F2 = − 6L3

[L

(−1

6wL3

)− 2

(− 1

24wL4

)]− wL = −wL

2 (23.23-d)

23.3.2.2 Concentrated Loads

30 For a concentrated load we can use the unit step function to find m(x). For a concentrated load Pacting at a from the left-hand end with b = L− a,

m(x) = −P (x− a)Ha gives m(L) = −Pbf(x) = − 1

2P (x− a)2Ha f(L) = − 12Pb

2

g(x) = − 16P (x− a)3Ha g(L) = − 1

6Pb3

(23.24)

where we define Ha = 0 if x < a, and Ha = 1 if x ≥ a.31 and

q = P (23.25-a)

MF1 =

2L2

[L

(−1

2Pb2

)− 3

(−1

6Pb3

)]== −Pb2a

L2 (23.25-b)



MF2 = − 1

L2

[L2 (−Pb)− 4L

(−1

2Pb2

)+ 6

(−1

6Pb3

)]=Pb

12(L2 − 2Lb+ b2

)(23.25-c)

= Pba2

L2 (23.25-d)

V F1 =6L3

[L

(−1

2Pb2

)− 2

(−1

6Pb3

)]= −Pb

2

L3(3L− 2b) = −Pb2

L3 (3a+ b) (23.25-e)

V F2 = −(

6L3

[L

(−1

2Pb2

)− 2

(−1

6Pb3

)]+ P

)= −Pa2

L3 (a+ 3b) (23.25-f)

32 If the load is applied at midspan (a = B = L/2), then the previous equation reduces to

MF1 = −PL

8(23.26)

MF2 =

PL

8(23.27)

V F1 = −P2

(23.28)

V F2 = −P2

(23.29)

23.4 Slope Deflection; Direct Solution

23.4.1 Slope Deflection Equations

33 In Eq. 23.10 and 23.11 if we let ∆ = v2 − v1 (relative displacement), ψ = ∆/L (rotation of the chordof the member), and K = I/L (stiffness factor) then the end equations are:

M1 = 2EK(2θ1 + θ2 − 3ψ) +MF1 (23.30)

M2 = 2EK(θ1 + 2θ2 − 3ψ) +MF2 (23.31)

34 Note that ψ will be positive if counterclockwise, negative otherwise.

35 From Eq. 23.30 and 23.31, we note that if a node has a displacement ∆, then both moments in theadjacent element will have the same sign. However, the moments in elements on each side of the nodewill have different sign.

23.4.2 Procedure

36 To illustrate the general procedure, we consider the two span beam in Fig. 23.5 under the appliedload, we will have three rotations θ1, θ2, and θ3 (i.e. three degrees of freedom) at the supports. Separatingthe spans from the supports, we can write the following equilibrium equations for each support

M12 = 0 (23.32-a)M21 +M23 = 0 (23.32-b)

M32 = 0 (23.32-c)

37 The three equilibrium equations in turn can be expressed in terms of the three unknown rotations,thus we can analyze this structure (note that in the slope deflection this will always be the case).

38 Using equations 23.30 and 23.31 we obtain

M12 = 2EK12(2θ1 + θ2) +MF12 (23.33-a)

M21 = 2EK12(θ1 + 2θ2) +MF21 (23.33-b)

M23 = 2EK23(2θ2 + θ3) (23.33-c)M32 = 2EK23(θ2 + 2θ3) (23.33-d)


Draft23.4 Slope Deflection; Direct Solution 23–9

��

��

��

a

M M M M

LL

12 21 23 32

1 2

P

1 2 3

P

1 2 3

Figure 23.5: Illustrative Example for the Slope Deflection Method

39 Substituting into the equations of equilibrium, we obtain

2 1 0K12 2(K12 +K23) K23

0 1 2

︸︷︷︸Stiffness Matrix

θ1θ2θ3

=

− MF

122EK12

−MF21

2E0

(23.34)

Where the fixed end moment can be separately determined.

40 Once the rotations are determined, we can then determine the moments from the slope deflectionequation Eq. 23.30.

41 The computational requirements of this method are far less than the one involved in the flexibilitymethod (or method of consistent deformation).

23.4.3 Algorithm

42 Application of the slope deflection method requires the following steps:

1. Sketch the deflected shape.

2. Identify all the unknown support degrees of freedom (rotations and deflections).

3. Write the equilibrium equations at all the supports in terms of the end moments.

4. Express the end moments in terms of the support rotations, deflections and fixed end moments.

5. Substitute the expressions obtained in the previous step in the equilibrium equations.

6. Solve the equilibrium equations to determine the unknown support rotation and/or deflections.

7. Use the slope deflection equations to determine the end moments.

8. Draw the moment diagram, careful about the difference in sign convention between the slopedeflection moments and the moment diagram.



23.4.4 Examples

Example 23-1: Propped Cantilever Beam, (Arbabi 1991)

Find the end moments for the beam of Fig. 23.6

��

��

��

10 m 5 m1 3

20 kN

2

Figure 23.6: Slope Deflection; Propped Cantilever Beam

Solution:

1. The beam is kinematically indeterminate to the third degree (θ2, ∆3, θ3), however by replacing thethe overhang by a fixed end moment equal to 100 kN.m at support 2, we reduce the degree of kinematicindeterminancy to one (θ2).2. The equilbrium relation is

M21 − 100 = 0 (23.35)

3. The members end moments in terms of the rotations are (Eq. 23.30 and 23.31)

M12 = 2EK12 (2θ1 + θ2) =210EIθ2 (23.36-a)

M21 = 2EK12 (θ1 + 2θ2) =410EIθ2 (23.36-b)

4. Substituting into the equilibrium equations

θ2 =10

4EIM21 =

250EI

(23.37)

or

M12 =210EIθ2 =

210EI

104EI

M21 =(2)EI(50)(10)EI

= 50 kN.m (23.38)

Example 23-2: Two-Span Beam, Slope Deflection, (Arbabi 1991)

Draw the moment diagram for the two span beam shown in Fig. 23.8 Solution:

1. The unknowns are θ1, θ2, and θ32. The equilibrium relations are

M21 +M23 = 0 (23.39-a)M32 = 0 (23.39-b)



��

��

��

��

20’ 15’ 15’1

3

79.5240.97

0

5 kips2 kip/ft

2 3

θ2 θ

Figure 23.7: Two-Span Beam, Slope Deflection

3. The fixed end moments are given by Eq. 23.23-b and 23.26

MF12 = −MF

21 = −wL2

12= − (−2)(20)2

12= 66.67 k.ft (23.40-a)

MF23 = −MF

32 = −PL8

= − (−5)(30)8

= 18.75 k.ft (23.40-b)


M12 = 2EK12(θ2) +MF12 =

2EIL1

θ2 +MF12 =

EI

10θ2 + 66.67 (23.41-a)

M21 = 2EK12(2θ2) +MF21 =

4EIL1

θ2 +MF21 =

EI

5θ2 − 66.67 (23.41-b)

M23 = 2EK23(2θ2 + θ3) +MF23 =

2EIL2

(2θ2 + θ3) +MF23

=EI

7.5θ2 +

EI

15θ3 + 18.75 (23.41-c)

M32 = 2EK23(θ2 + 2θ3) +MF32 =

2EIL2

(θ2 + 2θ3) +MF32

=EI

15θ2 +

EI

7.5θ3 − 18.75 (23.41-d)

(23.41-e)


EI

5θ2 − 66.67 +

EI

7.5θ2 +

EI

15θ3 + 18.75 = 0 (23.42-a)

EI

15θ2 +

EI

7.5θ3 − 18.75 = 0 (23.42-b)

or

EI

[5 11 2

]︸︷︷︸

Stiffness Matrix

{θ2θ3

}={

718.8281.25

}(23.43)



which will give EIθ2 = 128.48 and EIθ3 = 76.386. Substituting for the moments

M12 = 12.85 + 66.67 = 79.52 k.ft (23.44-a)

M21 =128.48

5− 66.67 = −40.97 k.ft (23.44-b)

M23 =128.48

7.5+

76.3815

+ 18.75 = 40.97 k.ft√

(23.44-c)

M32 =128.48

15+

76.387.5

− 18.75 = 0 k.ft√

(23.44-d)

43 The final moment diagram is also shown in Fig. 23.8. We note that the midspan moment has to beseparately computed from the equations of equilibrium in order to complete the diagram.

Example 23-3: Two-Span Beam, Slope Deflection, Initial Deflection, (Arbabi 1991)

Determine the end moments for the previous problem if the middle support settles by 6 inches, Fig.23.8.

��

��

��

��

20’ 15’ 15’

36"

21

Figure 23.8: Two Span Beam, Slope Deflection, Moment Diagram

Solution:

1. Since we are performing a linear elastic analysis, we can separately analyze the beam for supportsettlement, and then add then add the moments to those due to the applied loads.2. The unknowns are θ1, θ2, and θ33. The equilibrium relations are

M21 +M23 = 0 (23.45-a)M32 = 0 (23.45-b)


M12 = 2EK12

(θ2 − 3

∆L12

)=EI

10θ2 +

3EI400

(23.46-a)

M21 = 2EK12

(2θ2 − 3

∆L12

)=EI

5θ2 +

3EI400

(23.46-b)

M23 = 2EK23

(2θ2 + θ3 − 3

∆L23

)=EI

7.5θ2 +

EI

15θ3 +

EI

300(23.46-c)

M32 = 2EK23

(θ2 + 2θ3 − 3

∆L23

)=EI

15θ2 +

EI

7.5θ3 +

EI

300(23.46-d)




EI

5EIθ2 +

3EI400

+EI

15θ3 +

EI

300= 0 (23.47-a)

EI

15θ2 +

EI

7.5θ3 +

5EI300

= 0 (23.47-b)

or

EI

[100 2020 40

]︸︷︷︸Stiffness Matrix

{θ2θ3

}= EI

{ − 134−1

}(23.48)

which will give θ2 = − 5.5180 = −0.031 radians and θ3 = −1+ 5.5

940 = −0.0097 radians

6. Thus the additional moments due to the settlement are

M12 =EI

10(−0.031) +

3EI400

= 0.0044EI (23.49-a)

M21 =EI

5(−0.031) +

3EI400

= 0.0013EI (23.49-b)

M23 =EI

7.5(−0.031) +

EI

15(0.0097) +

EI

300= 0.0015EI (23.49-c)

M32 =EI

15θ2 +

EI

7.5(0.0097) +

EI

300= 0.

√(23.49-d)

Example 23-4: dagger Frames, Slope Deflection, (Arbabi 1991)

Determine the end moments for the frame shown in Fig. 23.9.

20’ 6’

5’

5’

10 kips

3 kips/ft∆ ∆

θ θ

θ

22

2

3

41

23 5

4

Figure 23.9: Frame Analysis by the Slope Deflection Method

Solution:

1. The effect of the 35 cantilever can be included by replacing it with its end moment.

M3 = −wLL2

= −(3)(6)(3) = −54 k.ft (23.50)



2. The unknowns displacements and rotations are

• ∆2 and θ2 at joint 2.

• θ3 and θ4 at joints 3 and 4.

We observe that due to the lack of symmetry, there will be a lateral displacement in the frame, andneglecting axial deformations, ∆2 = ∆3.3. The equilibrium relations are

M21 +M23 = 0 (23.51-a)M32 +M34 = −54 (23.51-b)

M43 = 0 (23.51-c)V12 + V43 − 10 = 0 (23.51-d)

Thus we have four unknown displacements and four equations. However, the last two equations are interms of the shear forces, and we need to have them in term of the end moments, this can be achievedthrough the following equilibrium relations

V12 =M12 +M21 + 50

L12(23.52-a)

V43 =M34 +M43

L34(23.52-b)

Hence, all four equations are now in terms of the moments.4. The fixed end moments fro member 23 are

MF21 = −PL

8= −−(10)(10)

8= 12.5 k.ft (23.53-a)

MF23 = −wL

2

12= −−(3)(20)2

12= 100 k.ft (23.53-b)


M12 = 2EK12

(θ2 − 3

∆2

L12

)+MF

12 = 0.2EI(θ2 − 0.3∆2) + 12.5 (23.54-a)

M21 = 2EK12

(2θ2 − 3

∆2

L21

)+MF

21 = 0.2EI(2θ2 − 0.3∆2)− 12.5 (23.54-b)

M23 = 2EK23 (2θ2 + θ3) +MF23 = 0.1EI(2θ2 + θ3) + 100. (23.54-c)

M32 = 2EK32 (θ2 + 2θ3) +MF32 = 0.1EI(θ2 + 2θ3)− 100. (23.54-d)

M34 = 2EK34

(2θ3 + θ4 − 3∆2

L34

)= 0.2EI(2θ3 + θ4 − 0.3∆2) (23.54-e)

M43 = 2EK43

(θ3 + 2θ4 − 3∆2

L34

)= 0.2EI(θ3 + 2θ4 − 0.3∆2) (23.54-f)

6. Substituting into the equilibrium equations and dividing by EI

6θ2 + θ3 − 0.6∆2 = −875EI

(23.55-a)

θ2 + 6θ3 + 2θ4 − 0.6∆2 =460EI

(23.55-b)

θ3 + 2θ4 − 0.3∆2 = 0 (23.55-c)

and the last equilibrium equation is obtained by substituting V12 and V43 and multiplying by 10/EI:

θ2 + θ3 + θ4 − 0.4∆2 = −83.3EI

(23.56)


Draft23.5 Moment Distribution; Indirect Solution 23–15

or

EI

6 1 0 −0.61 6 2 −0.60 1 2 −0.31 1 1 −0.4

︸︷︷︸Stiffness Matrix

θ2θ3θ4∆2

=

1EI

−8754600

−83.3

(23.57)

which will give

θ2θ3θ4∆2

=

1EI

−294.868.4−240.6−1, 375.7

(23.58)

7. Substitution into the slope deflection equations gives the end-moments

M12

M21

M23

M32

M34

M43

=

36.0−47.8847.88−115.8061.78

0

(23.59)

23.5 Moment Distribution; Indirect Solution

23.5.1 Background

43 The moment distribution is essentially a variation of the slope deflection method, however rather thansolving a system of n linear equations directly, the solution is achieved iteratively through a successiveseries of operations.

44 The method starts by locking all the joints, and then unlock each joint in succession, the internalmoments are then “distributed” and balanced until all the joints have rotated to their final (or nearlyfinal) position.

45 In order to better understand the method, some key terms must first be defined.

23.5.1.1 Sign Convention

46 The sign convention is the same as the one adopted for the slope deflection method, counter-clockwisemoment atelement’s end is positive.

23.5.1.2 Fixed-End Moments

47 Again fixed end moments are the same set of forces defined in the slope deflection method for a beamwhich is rigidly connected at both ends.

48 Consistent with the sign convention, the fixed end moments are the moments caused by the appliedload at the end of the beam (assuming it is rigidly connected).

23.5.1.3 Stiffness Factor

49 We define the stiffness factor as the moment required to rotate the end of a beam by a unit angle ofone radian, while the other end is fixed. From Eq. 23.10, we set θ2 = v1 = v2 = 0, and θ1 = 1, this will



yield

K =4EIL

Far End Fixed (23.60)

50 We note that this is slightly different than the definition given in the slope deflection method (I/L).

51 If the far end of the beam is hinged rather than fixed, then we will have a reduced stiffness factor.From Eq. 23.10 and 23.11, with M2 = v1 = v2 = 0, we obtain

M2 =2EIzL

(θ1 + 2θ2) = 0 (23.61-a)

⇒ θ2 = −θ12

(23.61-b)

Substituting into M1

M1 = 2EIzL (2θ1 + θ2)

θ2 = − θ12

}M1 = 3EI

L θ1 Far End Pinned (23.62)

52 Comparing this reduced stiffness 3EIL with the stiffness of a beam, we define the reduced stiffness

factor

Reduced Stiffness Factor =Kred

K=

34

Far End Pinned (23.63)

23.5.1.4 Distribution Factor (DF)

53 If a member is applied to a fixed-connection joint where there is a total of n members, then fromequilibrium:

M =M1 +M2 + · · ·+Mn (23.64)

However, from Eq. 23.30, and assuming the other end of the member to be fixed, then

M = K1θ +K2θ + · · ·+Knθ (23.65)

or

DFi =Mi

M=Ki

ΣKi

(23.66)

54 Hence if a moment M is applied at a joint, then the portion of M carried by a member connectedto this joint is proportional to the distribution factor. The stiffer the member, the greater the momentcarried.

55 Similarly, DF = 0 for a fixed end, and DF = 1 for a pin support.

23.5.1.5 Carry-Over Factor

56 Again from Eq. 23.30, and 23.31

M1 = 2EK(2θ1 + θ2 − 3ψ) +MF1 (23.67-a)

M2 = 2EK(θ1 + 2θ2 − 3ψ) +MF2 (23.67-b)

we observe that if one end of the beam is restrained (θ2 = ψ = 0), and there is no member load, thenthe previous equations reduce to

M1 = 2EK(2θ1)M2 = 2EK(θ1)

}⇒M2 =

12M1 (23.68)



57 Hence in this case the carry-over factor represents the fraction of M that is “carried over” from therotating end to the fixed one.

CO =12

Far End Fixed (23.69)

58 If the far end is pinned there is no carry over.

23.5.2 Procedure

59 The general procedure of the Moment Distribution method can be described as follows:

1. Constrain all the rotations and translations.

2. Apply the load, and determine the fixed end moments (which may be caused by element loading,or support translation).

3. At any given joint i equilibrium is not satisfied MFleft �=MF

right, and the net moment is Mi

4. We enforce equilibrium by applying at the node −Mi, in other words we balance the forces at thenode.

5. How much of Mi goes to each of the elements connected to node i depends on the distributionfactor.

6. But by applying a portion of −Mi to the end of a beam, while the other is still constrained, fromEq. 23.30, half of that moment must also be carried over to the other end.

7. We then lock node i, and move on to node j where these operations are repeated

(a) Sum moments

(b) Balance moments

(c) Distribute moments (K,DF )

(d) Carry over moments (CO)

(e) lock node

8. repeat the above operations until all nodes are balanced, then sum all moments.

9. The preceding operations can be easily carried out through a proper tabulation.

60 If an end node is hinged, then we can use the reduced stiffness factor and we will not carry overmoments to it.

61 Analysis of frame with unsymmetric loading, will result in lateral displacements, and a two stepanalysis must be performed (see below).

23.5.3 Algorithm

1. Calculate the stiffness (K = 4EI/L, however this can often be simplified to I/L) factor for all themembers and the distribution factors at all the joints.

2. If a member AB is pinned at B, then KAB = 3EI/L, and KBA = 4EI/L. Thus, we must applythe reduced stiffness factor to KAB only and not to KBA.

3. The carry-over factor is 12 for members with constant cross-section.

4. Find the fixed-end moments for all the members. Note that even if the end of a member is pinned,determine the fixed end moments as if it was fixed.

5. Start out by fixing all the joints, and release them one at a time.



6. If a node is pinned, start by balancing this particular node. If no node is pinned, start from eitherend of the structure.

7. Distribute the unbalanced moment at the released joint

8. Carry over the moments to the far ends of the members (unless it is pinned).

9. Fix the joint, and release the next one.

10. Continue releasing joints until the distributed moments are insignificant. If the last momentscarried over are small and cannot be distributed, it is better to discard them so that the jointsremain in equilibrium.

11. Sum up the moments at each end of the members to obtain the final moments.

23.5.4 Examples

Example 23-5: Continuous Beam, (Kinney 1957)

Solve for moments at A and B by moment distribution, using (a) the ordinary method, and (b) thesimplified method.

��

��

��

10k

12’ 10’ 10’

A

B C

25.812.90

3.22

9.5112.9

25.8

37.1

M

6.293.22 3.71

Solution:



1. For this example the fixed-end moments are computed as follows:

MFBC =

PL

8=

(10)(20)8

= +25.0 k.ft (23.70-a)

MFCB = −25.0 k.ft (23.70-b)

2. Since the relative stiffness is given in each span, the distribution factors are

DFAB = KAB

ΣK =5

∞+ 5= 0, (23.71-a)

DFBA = KBA

ΣK =58

= 0.625, (23.71-b)

DFBC = KBC

ΣK =38

= 0.375, (23.71-c)

DFCB = KCB

ΣK =33

= 1. (23.71-d)

3. The balancing computations are shown below.

Joint A B C Balance CO

Member AB BA BC CBK 5 5 3 3DF 0 0.625 0.375 1

FEM +25.0 -25.0❄

✑✑✰+12.5

❄✛ +25.0 C BC

-11.7✛ -23.4 -14.1 ✲ -7.0❄

B AB; CB

✑✑✰+3.5

❄✛ +7.0 C BC

-1.1 ✛ -2.2 -1.3 ✲ -0.6❄

B AB; CB

✑✑✰+0.3

❄✛ +0.6 C BC

-0.1 ✛ -0.2 -0.1 B ABTotal -12.9 -25.8 +25.8 0

4. The above solution is that referred to as the ordinary method, so named to designate the manner ofhandling the balancing at the simple support at C. It is known, of course, that the final moment mustbe zero at this support because it is simple.5. Consequently, the first step is to balance the fixed-end moment at C to zero. The carry-over isthen made immediately to B. When B is balanced, however, a carry-over must be made back to Csimply because the relative stiffness of BC is based on end C of this span being fixed. It is apparent,however, that the moment carried back to C (in this case, -7.0) cannot exist at this joint. Accordingly,it is immediately balanced out, and a carry-over is again made to B, this carry-over being considerablysmaller than the first. Now B is again balanced, and the process continues until the numbers involvedbecome too small to have any practical value.6. Alternatively, we can use the simplified method. It was previously shown that if the support at C issimple and a moment is applied at B, then the resistance of the span BC to this moment is reduced tothree-fourths of the value it would have had with C fixed. Consequently, if the relative stiffness of spanBC is reduced to three-fourths of the value given, it will not be necessary to carry over to C.

Joint A B C Balance CO

Member AB BA BC CBK 5 5 3

4×3 = 2.25 3.00DF 0 0.69 0.31 1

FEM +25.0 -25.0❄

✑✑✰+12.5

❄✛ +25.0 C BC

-12.9✛ -25.8 -11.7 B ABTotal -12.9 -25.8 +25.8 0



7. From the standpoint of work involved, the advantage of the simplified method is obvious. It shouldalways be used when the external (terminal) end of a member rests on a simple support, but it doesnot apply when a structure is continuous at a simple support. Attention is called to the fact that whenthe opposite end of the member is simply supported, the reduction factor for stiffness is always 3

4 for aprismatic member but a variable quantity for nonprismatic members.8. One valuable feature of the tabular arrangement is that of dropping down one line for each balancingoperation and making the carry-over on the same line. This practice clearly indicates the order ofbalancing the joints, which in turn makes it possible to check back in the event of error. Moreover, theplacing of the carry-over on the same line with the balancing moments definitely decreases the chanceof omitting a carry-over.9. The correctness of the answers may in a sense be checked by verifying that ΣM = 0 at each joint.However, even though the final answers satisfy this equation at every joint, this in no way a check onthe initial fixed-end moments. These fixed-end moments, therefore, should be checked with great carebefore beginning the balancing operation. Moreover, it occasionally happens that compensating errorsare made in the balancing, and these errors will not be apparent when checking ΣM = 0 at each joint.10. To draw the final shear and moment diagram, we start by drawing the free body diagram of eachbeam segment with the computed moments, and then solve from statics for the reactions:

12, 9 + 25.8− 12VA = 0 ⇒ VA = RA = 3.22 k ❄ (23.72-a)VA + V LB = 0 ⇒ V LB = 3.22 k ✻ (23.72-b)

25.8 + (10)(10)− 20V RB = 0 ⇒ V RB = 6.29 k ✻ (23.72-c)6.29 + VC − 10 = 0 ⇒ VC = RC3.71 k ✻ (23.72-d)

−V LB − V RB +RB = 0 ⇒ RB = 9.51 k ✻ (23.72-e)Check: RA +RB +RC − 10 = −3.22 + 9.51 + 3.71− 10 = 0

√(23.72-f)

M+BC = (3.71)(10) = 37.1 k.ft (23.72-g)

Example 23-6: Continuous Beam, Simplified Method, (Kinney 1957)

Using the simplified method of moment distribution, find the moments in the following continuousbeam. The values of I as indicated by the various values of K, are different for the various spans.Determine the values of reactions, draw the shear and bending moment diagrams, and sketch the deflectedstructure.

Solution:

1. Fixed-end moments:MFAO = −(0.5)(10) = −5.0 k.ft (23.73)

For the 1 k load:

MFAB =

Pab2

L2=

(1)(5)(152)202

= +2.8 k.ft (23.74-a)

MFBA =

Pa2b

L2=

(1)(52)(15)202

= −0.9 k.ft (23.74-b)



For the 4 k load:

MFAB = PL

8 = (4)(20)8 = +10.0 k.ft (23.75-a)

MFBA = −10.0 k.ft (23.75-b)

For the uniform load:

MFCD =

wL2

12=

(0.2)(152)12

= +3.8 k.ft (23.76-a)

MFDC = −3.8 k.ft (23.76-b)

2. The balancing operation is shown below

Joint A B C D Balance CO

Member AO AB BA BC CB CD DCK 0 20 3

4 (20) = 15 60 60 40 40DF 0 1 0.2 0.8 0.6 0.4 0

FEM -5.0 +12.8 -10.9 +3.8 -3.8-7.8 ✲ -3.9

❄◗◗�A BA

+2.9 +11.9 ✲+5.9❄◗◗�

B CB

✑✑✰-2.9

❄✛ -5.8 -3.9 ✲-1.9 C DC; BC

+0.6 +2.3 ✲+1.1❄◗◗�

B CB

✑✑✰-0.3

❄✛ -0.7 -0.4 ✲-0.2 C DC; BC

+0.1 +0.2 BTotal -5.0 +5.0 -11.2 +11.2 +0.5 -0.5 -5.9

3. The only new point in this example is the method of handling the overhanging end. It is obvious thatthe final internal moment at A must be 5.0 k.ft and, accordingly, the first step is to balance out 7.8 k.ft

of the fixed-end moment at AB, leaving the required 5.0 k.ft for the internal moment at AB. Since therelative stiffness of BA has been reduced to three-fourths of its original value, to permit considering thesupport at A as simple in the balancing, no carry-over from B to A is required.4. The easiest way to determine the reactions is to consider each span as a free body. End shears arefirst determined as caused by the loads alone on each span and, following this, the end shears caused bythe end moments are computed. These two shears are added algebraically to obtain the net end shearfor each span. An algebraic summation of the end shears at any support will give the total reaction.



Example 23-7: Continuous Beam, Initial Settlement, (Kinney 1957)

For the following beam find the moments at A,B, and C by moment distribution. The support atC settles by 0.1 in. Use E = 30, 000 k/in2.

Solution:

1. Fixed-end moments: Uniform load:

MFAB =

wL2

12=

(5)(202)12

= +167 k.ft (23.77-a)

MFBA = −167 k.ft (23.77-b)

Concentrated load:

MFCD =

PL

8=

(10)(30)8

= +37.5 k.ft (23.78-a)

MFDC = −37.5 k.ft (23.78-b)



Moments caused by deflection:

MFBC =

6EI∆L2

=6(30, 000)(1, 200)(0.1)

(120)2= +1, 500 k.in = +125 k.ft (23.79-a)

MFCB = +125 k.ft (23.79-b)

MFCD =

6EI∆L2

=(6)(30, 000)(7, 200)(0.1)

(360)2= 1, 000 k.in = −83 k.ft (23.79-c)

MFDC = −83 k.ft (23.79-d)

2. Moment distribution


Member AB BA BC CB CD DCK 10 10 10 10 3

4 (20) = 15 20DF 0 0.5 0.5 0.4 0.6 1

FEM Load +167 -167 +38 -38FEM ∆ +125 +125 -83 -83

✑✑✰+60

❄✛ +121 D CD

✑✑✰-28❄✛ -56 -84 C BC

+17✛ +35 +35 ✲❄+17◗◗�

B AB; CB

✑✑✰-3❄✛ -7 -10 C BC

+1 ✛ +2 +1 BTotal +185 -130 +130 +79 -79 0

The fixed-end moments caused by a settlement of supports have the same sign at both ends of eachspan adjacent to the settling support. The above computations have been carried to the nearest k.ft,which for moments of the magnitudes involved, would be sufficiently close for purposes of design.

Example 23-8: Frame, (Kinney 1957)

Find all moments by moment distribution for the following frame Draw the bending moment diagramand the deflected structure.

Solution:



1. The first step is to perform the usual moment distribution. The reader should fully understandthat this balancing operation adjusts the internal moments at the ends of the members by a series ofcorrections as the joints are considered to rotate, until ΣM = 0 at each joint. The reader should alsorealize that during this balancing operation no translation of any joint is permitted.2. The fixed-end moments are

MFBC =

(18)(12)(62)182

= +24 k.ft (23.80-a)

MFCB =

(18)(6)(122)182

= −48 k.ft (23.80-b)

3. Moment distribution


Member AB BA BC CB CD DCK 10 10 20 20 15 15DF 0 0.333 0.667 0.571 0.429 0

FEM +24.0 -48.0 FEM

✑✑✰+13.7

❄✛ +27.4 +20.6 ✲+10.3 C DC; BC

-6.3✛ -12.6 -25.1 ✲❄-12.5◗◗�

B AB; CB

✑✑✰+3.6

❄✛ +7.1 +5.4 ✲ +2.7 C BC; DC

-0.6✛ -1.2 -2.4 ✲❄-1.2◗◗�

B AB; CB

✑✑✰+0.3

❄✛ +0.7 +0.5 ✲ +0.02 C BC; DC

-0.1 -0.2 BTotal -6.9 -13.9 +13.9 -26.5 +26.5 +13.2

4. The final moments listed in the table are correct only if there is no translation of any joint. It istherefore necessary to determine whether or not, with the above moments existing, there is any tendencyfor side lurch of the top of the frame.5. If the frame is divided into three free bodies, the result will be as shown below.

Inspection of this sketch indicates that if the moments of the first balance exist in the frame, there is anet force of 1.53 − 0.80 = 0.73 k tending to sway the frame to the left. In order to prevent side-sway,and thus allow these moments to exist (temporarily, for the purpose of the analysis), it is necessary thatan imaginary horizontal force be considered to act to the right at B or C. This force is designated asthe artificial joint restraint (abbreviated as AJR) and is shown below.

6. This illustration now shows the complete load system which would have to act on the structure if thefinal moments of the first balance are to be correct. The AJR, however, cannot be permitted to remain,



and thus its effect must be cancelled. This may be accomplished by finding the moments in the frameresulting from a force equal but opposite to the AJR and applied at the top.7. Although it is not possible to make a direct solution for the moments resulting from this force, theymay be determined indirectly. Assume that some unknown force P acts on the frame, as shown below

and causes it to deflect laterally to the left, without joint rotation, through some distance ∆. Now,regardless of the value of P and the value of the resulting ∆, the fixed-end moments induced in the endsof the columns must be proportional to the respective values of KM .

62 Recalling that the fixed end moment is MF = 6EI ∆L2 = 6EKm∆, where Km = I

L2 = KL we can write

∆ =MFAB

6EKm=MFDC

6EKm(23.81-a)

⇒ MFAB

MFDC

=KABm

KDCm

=1015

(23.81-b)

63 These fixed-end moments could, for example, have the values of −10 and −15 k.ft or −20 and −30, or−30 and −45, or any other combination so long as the above ratio is maintained. The proper procedureis to choose values for these fixed-end moments of approximately the same order of magnitude as theoriginal fixed-end moments due to the real loads. This will result in the same accuracy for the resultsof the balance for the side-sway correction that was realized in the first balance for the real loads.Accordingly, it will be assumed that P , and the resulting ∆, are of such magnitudes as to result in thefixed-end moments shown below

8. Obviously, ΣM = 0 is not satisfied for joints B and C in this deflected frame. Therefore these jointsmust rotate until equilibrium is reached. The effect of this rotation is determined in the distributionbelow




Member AB BA BC CB CD DCK 10 10 20 20 15 15DF 0 0.333 0.667 0.571 0.429 0

FEM -30.0 -30.0 ✑✑✰-45.0

❄-45.0

✑✑✰+12.9

❄✛ +25.8 +19.2 ✲+9.6 C BC; DC

+2.8✛ +5.7 +11.4 ✲❄+5.7◗◗�

B AB; CB

✑✑✰-1.6❄✛ -3.3 -2.4 ✲ -1.2 C BC; DC

+0.2✛ +0.5 +1.1 ✲❄+0.5◗◗�

B AB; CB-0.3 -0.2 ✲ -0.1 C

Total -27.0 -23.8 +23.8 +28.4 -28.4 -36.7

9. During the rotation of joints B and C, as represented by the above distribution, the value of ∆ hasremained constant, with P varying in magnitude as required to maintain ∆.10. It is now possible to determine the final value of P simply by adding the shears in the columns.The shear in any member, without external loads applied along its length, is obtained by adding theend moments algebraically and dividing by the length of the member. The final value of P is the forcenecessary to maintain the deflection of the frame after the joints have rotated. In other words, it is theforce which will be consistent with the displacement and internal moments of the structure as determinedby the second balancing operation. Hence this final value of P will be called the consistent joint force(abbreviated as CJF).11. The consistent joint force is given by

CJF =+27.0 + 23.8

26+

28.4 + 36.726

= 1.95 + 2.50 = 4.45 k (23.82)

and inspection clearly indicates that the CJF must act to the left.12. Obviously, then, the results of the last balance above are moments which will exist in the frame whena force of 4.45 k acts to the left at the top level. It is necessary, however, to determine the momentsresulting from a force of 0.73 k acting to the left at the top level, and some as yet unknown factor “z”times 4.45 will be used to represent this force acting to the left.

13. The free body for the member BC is shown above. ΣH = 0 must be satisfied for this figure, and ifforces to the left are considered as positive, the result is 4.45z − 0.73 = 0, and

z = +0.164. (23.83)

If this factor z = +0.164 is applied to the moments obtained from the second balance, the result willbe the moments caused by a force of 0.73 k acting to the left at the top level. If these moments arethen added to the moments obtained from the first balance, the result will be the final moments for theframe, the effect of the AJR having been cancelled. This combination of moments is shown below.

Joint A B C DMember AB BA BC CB CD DC

M from 1st balance -6.9 -13.9 +13.9 -26.5 +26.5 +13.2z × M from 2nd balance -4.4 -3.9 +3.9 +4.7 -4.7 -6.0

Final moments -11.3 -17.8 +17.8 -21.8 +21.8 +7.2

14. If the final moments are correct, the shears in the two columns of the frame should be equal andopposite to satisfy ΣH = 0 for the entire frame. This check is expressed as

+11.3 + 17.826

+−21.8− 7.2

26= 0, (23.84)

and+1.12− 1.11 = 0(nearly) (23.85)



The signs of all moments taken from the previous table have been reversed to give the correct signsfor the end moments external to the columns. It will be remembered that the moments considered inmoment distribution are always internal for each member. However, the above check actually considerseach column as a free body and so external moments must be used.15. The moment under the 18 k load is obtained by treating BC as a free body:

M18 = (5.77)(12)− 17.8 = +51.5 k.ft (23.86)

16. The direction of side-lurch may be determined from the obvious fact that the frame will always lurchin a direction opposite to the AJR. If required, the magnitude of this side lurch may be found. Theprocedure which follows will apply.

A force P of sufficient magnitude to result in the indicated column moments and the lurch ∆ was appliedto the frame. During the second balance this value of ∆ was held constant as the joints B and C rotated,and the value of P was considered to vary as necessary. The final value of P was found to be 4.45 k. Since∆ was held constant, however, its magnitude may be determined from the equation M = 6EI∆/L2,where M is the fixed-end moment for either column, I is the moment of inertia of that column, and Lis the length. This ∆ will be the lurch for 4.45 k acting at the top level. For any other force actinghorizontally, ∆ would vary proportionally and thus the final lurch of the frame would be the factor zmultiplied by the ∆ determined above.

Example 23-9: Frame with Side Load, (Kinney 1957)

Find by moment distribution the moments in the following frame



Solution:

The first balance will give the results shown

AB BA BC CB CD DC-7.2 -14.6 +14.6 -22.5 -22.5 0

A check of the member BC as a free body for ΣH = 0 will indicate that an AJR is necessary as follows:

AJR+ 0.84− 0.87− 5.0 = 0 (23.87)

from whichAJR = +5.03 in the direction assumed (23.88)

The values of KM for the two columns are shown, with KM for column CD being K/2L because of thepin at the bottom. The horizontal displacement ∆ of the top of the frame is

assumed to cause the fixed-end moments shown there. These moments are proportional to the valuesof KM and of approximately the same order of magnitude as the original fixed-end moments due to thereal loads. The results of balancing out these moments are

AB BA BC CB CD DC-34.4 -28.4 +28.4 +23.6 -23.6 0

CJF =+34.4 + 28.4 + 23.6

26= 3.32 k (23.89)

and5.03− z(3.32) = 0, (23.90)

from whichz = +1.52. (23.91)

The final results are



AB BA BC CB CD DCM from 1st balance -7.2 -14.6 +14.6 -22.5 +22.5 0z × M 2nd balance -52.1 -43.0 +43.0 +35.8 -35.8 0Final moments -59.3 -57.6 +57.6 +13.3 -13.3 0

If these final moments are correct, the sum of the column shears will be 5.0 k: Sum of column shears:

ΣV =59.3 + 57.6 + 13.3

26= 5.01 k (23.92)

The 5 k horizontal load acting at C enters into the problem only in connection with the determinationof the AJR. If this load had been applied to the column CD between the ends, it would have resultedin initial fixed-end moments in CD and these would be computed in the usual way. In addition, such aload would have entered into the determination of the AJR, since the horizontal reaction of CD againstthe right end of BC would have been computed by treating CD as a free body.

Example 23-10: Moment Distribution on a Spread-Sheet

Analyse the following frame

12’ 8’

10’15’

A

B C

D

20 k

I = cst

Solution:



Sheet1

Artificially Restrained StructureV 6EI/L^2 M^F

A B C D H_A -2.84 AB 0.80 -8AB BA BC CB CD DC Balance CO H_D 6.826355685 DC 1.80 -18

Lenght 15 15 20 20 10 10 AJR 3.98 Delta -10EI 30 30 30 30 30 30 "arbitrary" Delta to the left, will cause -ve M^FK 2.0 2.0 1.5 1.5 3.0 3.0 Frame sways to the rightDF 1.0 0.6 0.4 0.3 0.7 1.0FEM 38.4 -57.6

-11.0 -21.9 -16.5 -8.2 B AB, CB11.0 21.9 43.9 21.9 C BC, DC

-3.1 -6.3 -4.7 -2.4 B AB, CB0.4 0.8 1.6 0.8 C BC, DC

-0.1 -0.2 -0.2 -0.1 B AB, CB0.0 0.0 0.1 0.0 C BC, DC

Total -14.2 -28.4 28.5 -45.5 45.5 22.8 .Horizontal displacement Delta

A B C DAB BA BC CB CD DC Balance CO

Lenght 15 15 20 20 10 10EI 30 30 30 30 30 30 "arbitrary"K 2.0 2.0 1.5 1.5 3.0 3.0DF 1.0 0.6 0.4 0.3 0.7 1.0FEM -8.0 -8.0 -18.0 -18.0

2.3 4.6 3.4 1.7 B AB, CB V2.7 5.4 10.9 5.4 C BC, DC H_A -0.77

-0.8 -1.6 -1.2 -0.6 B AB, CB H_D -1.90.1 0.2 0.4 0.2 C BC, DC CJF -2.68

-0.1 0.0 B AB, CB z 1.49Total -6.5 -5.0 5.0 6.8 -6.8 -12.4 .

Final MomentsAB BA BC CB CD DC

1st -14.2 -28.4 28.5 -45.5 45.5 22.8z 2nd -9.6 -7.5 7.5 10.0 -10.0 -18.4Final -23.9 -35.9 35.9 -35.5 35.5 4.4

Page 1


Draft

Chapter 24

REINFORCED CONCRETEBEAMS; Part I

24.1 Introduction

1 Recalling that concrete has a tensile strength (f ′t) about one tenth its compressive strength (f ′c),concrete by itself is a very poor material for flexural members.

2 To provide tensile resistance to concrete beams, a reinforcement must be added. Steel is almostuniversally used as reinforcement (longitudinal or as fibers), but in poorer countries other indigenousmaterials have been used (such as bamboos).

3 The following lectures will focus exclusively on the flexural design and analysis of reinforced con-crete rectangular sections. Other concerns, such as shear, torsion, cracking, and deflections are left forsubsequent ones.

4 Design of reinforced concrete structures is governed in most cases by the Building Code Requirementsfor Reinforced Concrete, of the American Concrete Institute (ACI-318). Some of the most relevantprovisions of this code are enclosed in this set of notes.

5 We will focus on determining the amount of flexural (that is longitudinal) reinforcement required at agiven section. For that section, the moment which should be considered for design is the one obtainedfrom the moment envelope at that particular point.

24.1.1 Notation

6 In R/C design, it is customary to use the following notation

As Area of steelb Widthc Distance from top of compressive fibers to neutral axisd Distance from the top of the compressive fibers to the centroid

of the reinforcementf ′c Concrete compressive strengthf ′r Concrete modulus of rupturef ′s Steel stressf ′t Concrete tensile strengthfy Steel yield stress (equivalent to Fy in AISC)h Heightρ Steel ratio, Asbd

Draft24–2 REINFORCED CONCRETE BEAMS; Part I

24.1.2 Modes of Failure

7 A reinforced concrete beam is a composite structure where concrete provides the compression and steelthe tension.

8 Failure is initiated by

1. Yielding of the steel when the steel stress reaches the yield stress (fs = fy). This occurs if wedo not have enough reinforcement that is the section is under-reinforced. This will result inexcessive rotation and deformation prior to failure.

2. Crushing of the concrete, when the concrete strain reaches its ultimate value (εc = εu = 0.003),ACI 318: 10.2.3. This occurs if there is too much reinforcement that is the section is over-reinforced. This is a sudden mode of failure.

9 Ideally in an optimal (i.e. most efficient use of materials) design, a section should be dimensionedsuch that crushing of concrete should occur simultaneously with steel yielding. This would then be abalanced design.

10 However since concrete crushing is a sudden mode of failure with no prior warning, whereas steelyielding is often accompanied by excessive deformation (thus providing ample warning of an imminentfailure), design codes require the section to be moderately under-reinforced.

24.1.3 Analysis vs Design

11 In R/C we always consider one of the following problems:

Analysis: Given a certain design, determine what is the maximum moment which can be applied.

Design: Given an external moment to be resisted, determine cross sectional dimensions (b and h) aswell as reinforcement (As). Note that in many cases the external dimensions of the beam (b andh) are fixed by the architect.

12 We often consider the maximum moment along a member, and design accordingly.

24.1.4 Basic Relations and Assumptions

13 In developing a design/analysis method for reinforced concrete, the following basic relations will beused:

1. Equilibrium: of forces and moment at the cross section. 1) ΣFx = 0 or Tension in the reinforcement= Compression in concrete; and 2) ΣM = 0 or external moment (that is the one obtained fromthe moment envelope) equal and opposite to the internal one (tension in steel and compression ofthe concrete).

2. Material Stress Strain: We recall that all normal strength concrete have a failure strain εu = .003in compression irrespective of f ′c.

14 Basic assumptions used:

Compatibility of Displacements: Perfect bond between steel and concrete (no slip). Note that thosetwo materials do also have very close coefficients of thermal expansion under normal temperature.

Plane section remain plane ⇒ strain is proportional to distance from neutral axis.


Draft24.2 Cracked Section, Ultimate Strength Design Method 24–3

24.1.5 ACI Code

15 The ACI code is based on limit strength, or ΦMn ≥ Mu thus a similar design philosophy is used asthe one adopted by the LRFD method of the AISC code, ACI-318: 8.1.1; 9.3.1; 9.3.2

16 The required strength is based on (ACI-318: 9.2)

U = 1.4D + 1.7L (24.1)= 0.75(1.4D+ 1.7L+ 1.7W ) (24.2)

17 We should consider the behaviors of a reinforced concrete section under increasing load:

1. Section uncracked

2. Section cracked, elastic

3. Section cracked, limit state

The second analysis gives rise to the Working Stress Design (WSD) method (to be covered in StructuralEngineering II), and the third one to the Ultimate Strength Design (USD) method.

24.2 Cracked Section, Ultimate Strength Design Method

24.2.1 Equivalent Stress Block

18 In determining the limit state moment of a cross section, we consider Fig. 24.1. Whereas the straindistribution is linear (ACI-318 10.2.2), the stress distribution is non-linear because the stress-straincurve of concrete is itself non-linear beyond 0.5f ′c.

19 Thus we have two alternatives to investigate the moment carrying capacity of the section, ACI-318:10.2.6

1. Use the non-linear stress distribution.

2. Use a simpler equivalent stress distribution.

20 The ACI code follows the second approach. Thus we seek an equivalent stress distribution suchthat:

1. The resultant force is equal

2. The location of the resultant is the same.

We note that this is similar to the approach followed in determining reactions in a beam subjected to adistributed load when the load is replaced by a single force placed at the centroid.

21 From the actual stress distribution we have the following relations

C = T = Asfs = AsfyM = T (d− βc)

22 We select an equivalent rectangular stress distribution, often referred to as the Whitney StressBlock such that the constant stress is equal to 0.85f ′c and is applied over a depth a = β1c, ACI-318:10.2.7

23 Comparing the actual and the equivalent stress distributions, the two requirements previously statedtranslate into the following mathematical relations

αf ′cbc = 0.85f ′cbβ1cβc = 1

2β1c



Figure 24.1: Cracked Section, Limit State

Those two equations yieldβ1 = 2βαβ1

= .85

24 We have two equations and three unknowns (α, β1, and β). Thus we need to use test data to solvethis problem1. From experimental tests, the following relations are obtained

f ′c (ppsi) < 4,000 5,000 6,000 7,000 8,000α .72 .68 .64 .60 .56β .425 .400 .375 .350 .325β1 .85 .80 .75 .70 .65

Thus we have a general equation for β1 (ACI-318 10.2.7.3):

β1 = .85 if f ′c ≤ 4, 000= .85− (.05)(f ′c − 4, 000) 1

1,000 if 4, 000 < f ′c < 8, 000(24.3)

Figure 24.2: Whitney Stress Block

1This approach is often used in Structural Engineering. Perform an analytical derivation, if the number of unknownsthen exceeds the number of equations, use experimental data.



24.2.2 Balanced Steel Ratio

25 Next we seek to determine the balanced steel ratio ρb such that failure occurs by simultaneousyielding of the steel fs = fy and crushing of the concrete εc = 0.003, ACI-318: 10.3.2 We willseparately consider the two failure possibilities:

Tension Failure: we stipulate that the steel stress is equal to fy:

ρ = Asbd

Asfy = .85f ′cab = .85f ′cbβ1c

}⇒ c =

ρfy0.85f ′cβ1

d (24.4)

Compression Failure: where the concrete strain is equal to the ultimate strain; From the straindiagram

εc = 0.003cd = .003

.003+εs

}⇒ c =

.003fsEs

+ .003d (24.5)

26 Balanced Design is obtained by equating Eq. 24.4 to Eq. 24.5 and by replacing ρ by ρb and fs by fy:

ρfy0.85f ′

cβ1d = .003

fsEs

+.003d

fs = fyρ = ρb

ρbfy.85f ′cβ1

d =.003

fyEs

+ .003d

When we replace Es by 29, 000 ksi we obtain

ρb = .85β1f ′cfy

87, 00087, 000 + fy

(24.6)

This ρb corresponds to the only combination of b, d and As which will result in simultaneous yielding ofthe steel and crushing of the concrete, that is an optimal design.

27 Because we need to have ample warning against failure, hence we prefer to have an under-reinforcedsection. Thus, the ACI code stipulates:

ρ < .75ρb (24.7)

28 In practice, depending on the relative cost of steel/concrete and of labour it is common to select lowervalues of ρ. If ρ < 0.5ρb (thus we will have a deeper section) then we need not check for deflection.

29 A minimum amount of reinforcement must always be used to prevent temperature and shrinkagecracks:

ρmin ≥ 200fy

(24.8)

30 The ACI code adopts the limit state design method

MD = ΦMn > Mu Φ = Φb = 0.9 (24.9)

24.2.3 Analysis

Given As, b, d, f ′c, and fy determine the design moment:

1. ρact = Asbd

2. ρb = (.85)β1f ′c

fy87

87+fy



3. If ρact < ρb (that is failure is triggered by yielding of the steel, fs = fy)

a = Asfy.85f ′

cbFrom Equilibrium

MD = ΦAsfy(d− a

2

)}MD = ΦAsfy

(d− 0.59

Asfyf ′cb

)︸︷︷︸

Mn

Combining this last equation with ρ = Asbd yields

MD = Φρfybd2(

1− .59ρfyf ′c

)(24.10)

4. † If ρact > ρb is not allowed by the code as this would be an over-reinforced section which wouldfail with no prior warning. However, if such a section exists, and we need to determine its momentcarrying capacity, then we have two unknowns:

(a) Steel strain εs (which was equal to εy in the previous case)

(b) Location of the neutral axis c.

We have two equations to solve this problem

Equilibrium: of forces

c =Asfs.85f ′cbβ1

(24.11)

Strain compatibility: since we know that at failure the maximum compressive strain εc is equalto 0.003. Thus from similar triangles we have

c

d=

.003.003 + εs

(24.12)

Those two equations can be solved by either one of two methods:

(a) Substitute into one single equation

(b) By iteration

Once c and fs = Eεs are determined then

MD = ΦAsfs

(d− β1c

2

)(24.13)

24.2.4 Design

31 We distinguish between two cases. The first one has dimensions as well as steel area unknown, thesecond has the dimensions known (usually specified by the architect or by other constraints), and weseek As.

b, d and As unknowns and MD known:

1. We assume that ρ = .75ρb with ρb determined from Eq. 24.6

ρ = 0.75[.85β1

f ′cfy

87, 00087, 000 + fy

]

Often times, a designer will take a smaller portion of ρb if the relative cost of steel is higherthan the one of concrete, resulting in deeper sections. The ratio may range from 0.4 to 0.75.



2. From Eq. 24.10

MD = Φ ρfy

(1− .59ρfy

f ′c

)︸︷︷︸

R

bd2

or

R = ρfy

(1− .59ρfy

f ′c

)(24.14)

which does not depend on unknown quantities2. Then solve for bd2:

bd2 =MD

ΦR(24.15)

Note that in most case R is around 1 ksi

3. solve for b and d (this will require either an assumption on one of the two, or on their ratio).

4. As = ρbd

dag b, d and Md known, As unknown: In this case there is no assurance that we can have a designwith ρb. If the section is too small, then it will require too much steel resulting in an over-reinforcedsection.

We will again have an iterative approach

1. Since we do not know if the steel will be yielding or not, use fs.

2. Assume an initial value for a (a good start is a = d5 )

3. Assume initially that fs = fy4. Check equilibrium of moments (ΣM = 0)

As =MD

Φfs(d− a

2

) (24.16)

5. Check equilibrium of forces in the x direction (ΣFx = 0)

a =Asfs.85f ′cb

(24.17)

6. Check assumption of fs from the strain diagram

εsd− c =

.003c⇒ fs = Es

d− cc.003 < fy (24.18)

where c = aβ1

.

7. Iterate until convergence is reached.

Example 24-1: Ultimate Strength Capacity

Determine the ultimate Strength of a beam with the following properties: b = 10 in, d = 23 in,As = 2.35 in2, f ′c = 4, 000 psi and fy = 60 ksi.Solution:

ρact = Asbd = 2.35

(10)(23) = .0102

ρb = .85β1f ′c

fy87

87+fy= (.85)(.85) 4

6087

87+60 = .02885 > ρact√

a = Asfy.85f ′

cb= (2.35)(60)

(.85)(4)(10) = 4.147 in

Mn = (2.35)(60)(23− 4.1472 ) = 2, 950 k.in

MD = ΦMn = (.9)(2, 950) = 2, 660 k.in

2Note analogy with Eq. 14.5 Mp = FyZ for stell beams.



Note that from the strain diagram

c =a

0.85=

4.4140.85

= 4.87 in

Alternative solution

Mn = ρactfybd2(1− .59ρact fyf ′

c

)= Asfyd

(1− .59ρact fyf ′

c

)= (2.35)(60)(23)

[1− (.59)604 (.01021)

]= 2, 950 k.in = 245 k.ft

Example 24-2: Beam Design I

Design a 15 ft beam to support a dead load of 1.27 k/ft, a live load of 2.44 k/ft using a 3,000 psiconcrete and 40 ksi steel. Neglect beam weightSolution:

wu = 1.4(1.27) + 1.7(2.44) = 5.92 k/ft

MD = (5.92) k/ft(15)2 ft2

8 (12) in/ft = 2, 000 k.in

ρb = .85β1f ′c

fy87

87+fy

= (.85)(.85) 340

8787+40 = 0.040

ρ = .75ρb = .75(0.04) = 0.030R = ρfy

(1− .59ρ fyf ′

c

)= (0.03)(40)

[1.− (0.59)(0.03)403

]= 0.917 ksi

bd2 = Md

ΦR = 2,000(0.9)(0.917)

k.in in2

k = 2, 423 in3

Assume b = 10 in, this will give d =√

2,42310 = 15.57 in. We thus adopt b = 10 in and d = 16 in.

Finally,As = ρbd = (0.030)(10)(16) = 4.80 in2

we select 3 bars No. 11

Example 24-3: Beam Design II

Select the reinforcement for a cross section with b = 11.5 in; d = 20 in to support a design momentMd = 1, 600 k.in using f ′c = 3 ksi; and fy = 40 ksi

Solution:

1. Assume a = d5 = 20

5 = 4 in and fs = fy

2. Equilibrium of moments:

As =MD

Φfy(d− a2 )

==1, 600 k.in

(.9)(40) ksi(20− 42 ) in

= 2.47 in2

3. Check equilibrium of forces:

a =Asfy.85f ′cb

=(2.47) in2(40) ksi

(.85)(3) ksi(11.5) in= 3.38 in


Draft24.3 ACI Code 24–9

4. We originally assumed a = 4, at the end of this first iteration a = 3.38, let us iterate again witha = 3.30

5. Equilibrium of moments:

As =MD

Φfy(d− a2 )

==1, 600 k.in

(.9)(40) ksi(20− 3.32 ) in

= 2.42 in2

6. Check equilibrium of forces:

a =Asfy.85f ′cb

=(2.42) in2(40) ksi

(.85)(3) ksi(11.5) in= 3.3 in

√

7. we have converged on a.

8. Actual ρ is ρact = 2.42(11.5)(20) = .011

9. ρb is equal to

ρb = .85β1f ′cfy

8787 + fy

= (.85)(.85)340

8787 + 40

= .037

10. ρmax = .75ρ = (0.75)(0.037) = .0278 > 0.011√

thus fs = fy.

24.3 ACI Code

Attached is an unauthorized copy of some of the most relevant ACI-318-89 design code provisions.8.1.1 - In design of reinforced concrete structures, members shall be proportioned for adequate

strength in accordance with provisions of this code, using load factors and strength reduction factors Φspecified in Chapter 9.

8.3.1 - All members of frames or continuous construction shall be designed for the maximum effectsof factored loads as determined by the theory of elastic analysis, except as modified according to Section8.4. Simplifying assumptions of Section 8.6 through 8.9 may be used.

8.5.1 - Modulus of elasticity Ec for concrete may be taken as W 1.5c 33

√f ′c ( psi) for values of Wc

between 90 and 155 lb per cu ft. For normal weight concrete, Ec may be taken as 57, 000√f ′c.

8.5.2 - Modulus of elasticity Es for non-prestressed reinforcement may be taken as 29,000 psi.9.1.1 - Structures and structural members shall be designed to have design strengths at all sections

at least equal to the required strengths calculated for the factored loads and forces in such combinationsas are stipulated in this code.

9.2 - Required Strength9.2.1 - Required strength U to resist dead load D and live load L shall be at least equal to

U = 1.4D+ 1.7L

9.2.2 - If resistance to structural effects of a specified wind load W are included in design, thefollowing combinations of D, L, and W shall be investigated to determine the greatest required strengthU

U = 0.75(1.4D+ 1.7L+ 1.7W )

where load combinations shall include both full value and zero value of L to determine the more severecondition, and

U = 0.9D+ 1.3W

but for any combination of D, L, and W, required strength U shall not be less than Eq. (9-1).9.3.1 - Design strength provided by a member, its connections to other members, and its cross

sections, in terms of flexure, axial load, shear, and torsion, shall be taken as the nominal strengthcalculated in accordance with requirements and assumptions of this code, multiplied by a strengthreduction factor Φ.



9.3.2 - Strength reduction factor Φ shall be as follows:9.3.2.1 - Flexure, without axial load 0.909.4 - Design strength for reinforcement Designs shall not be based on a yield strength of reinforcement

fy in excess of 80,000 psi, except for prestressing tendons.10.2.2 - Strain in reinforcement and concrete shall be assumed directly proportional to the distance

from the neutral axis, except, for deep flexural members with overall depth to clear span ratios greaterthan 2/5 for continuous spans and 4/5 for simple spans, a non-linear distribution of strain shall beconsidered. See Section 10.7.

10.2.3 - Maximum usable strain at extreme concrete compression fiber shall be assumed equal to0.003.

10.2.4 - Stress in reinforcement below specified yield strength fy for grade of reinforcement usedshall be taken as Es times steel strain. For strains greater than that corresponding to fy, stress inreinforcement shall be considered independent of strain and equal to fy.

10.2.5 - Tensile strength of concrete shall be neglected in flexural calculations of reinforced concrete,except when meeting requirements of Section 18.4.

10.2.6 - Relationship between concrete compressive stress distribution and concrete strain may beassumed to be rectangular, trapezoidal, parabolic, or any other shape that results in prediction ofstrength in substantial agreement with results of comprehensive tests.

10.2.7 - Requirements of Section 10.2.5 may be considered satisfied by an equivalent rectangularconcrete stress distribution defined by the following:

10.2.7.1 - Concrete stress of 0.85f ′c shall be assumed uniformly distributed over an equivalent com-pression zone bounded by edges of the cross section and a straight line located parallel to the neutralaxis at a distance (a = β1c) from the fiber of maximum compressive strain.

10.2.7.2 - Distance c from fiber of maximum strain to the neutral axis shall be measured in adirection perpendicular to that axis.

10.2.7.3 - Factor β1 shall be taken as 0.85 for concrete strengths f ′c up to and including 4,000 psi.For strengths above 4,000 psi, β1 shall be reduced continuously at a rate of 0.05 for each 1000 psi ofstrength in excess of 4,000 psi, but β1 shall not be taken less than 0.65.

10.3.2 - Balanced strain conditions exist at a cross section when tension reinforcement reaches thestrain corresponding to its specified yield strength fy just as concrete in compression reaches its assumedultimate strain of 0.003.

10.3.3 - For flexural members, and for members subject to combined flexure and compressive axialload when the design axial load strength (ΦPn) is less than the smaller of (0.10f ′cAg) or (ΦPb), the ratioof reinforcement p provided shall not exceed 0.75 of the ratio ρb that would produce balanced strainconditions for the section under flexure without axial load. For members with compression reinforcement,the portion of ρb equalized by compression reinforcement need not be reduced by the 0.75 factor.

10.3.4 - Compression reinforcement in conjunction with additional tension reinforcement may beused to increase the strength of flexural members.

10.5.1 - At any section of a flexural member, except as provided in Sections 10.5.2 and 10.5.3, wherepositive reinforcement is required by analysis, the ratio ρ provided shall not be less than that given by

ρmin =200fy


Draft

Chapter 25

REINFORCED CONCRETEBEAMS; Part II

25.1 T Beams, (ACI 8.10)

1 Equivalent width for uniform stress, Fig. 25.1 must satisfy the following requirements (ACI 8.10.2):

b

bw

hf

be

Figure 25.1: T Beams

Draft25–2 REINFORCED CONCRETE BEAMS; Part II

1. 12 (b− bw) ≤ 8hf

2. b < 4bw for isolated T beams only

3. hf > bw2

4. b < L4

2 Two possibilities:

1. Neutral axis within the flanges (c < hf ) ⇒ rectangular section of width b, Fig. 25.2.

2. Neutral axis in the web (c > hf ) ⇒ T beam.

b

Figure 25.2: T Beam as Rectangular Section

d

hf

b

c

εs

As fy

a=β1c

.85f’cεu

Figure 25.3: T Beam Strain and Stress Diagram

3 For T beams, we have a large concrete area, start by assuming that failure will occur by steel yielding,Fig. 25.3.

4 The approach consists in decomposing As into 2 components

1. Asf → resists compression force in (b − bw)hf

2. (As −Asf )→ resists compression force in bwc


Draft25.1 T Beams, (ACI 8.10) 25–3

25.1.1 Review

5 Given, b, d, hf , As, f ′c, fy, determine the moment capacity M , Fig. 25.4.

= +

As Asf As-Asf

Figure 25.4: Decomposition of Steel Reinforcement for T Beams

6 The moment is obtained fromFlanges:

Asf = .85f ′c(b−bw)hffy

ΣF = 0

Mn1 = Asffy(d− hf2 ) ΣM = 0

Web:

a = (As−Asf )fy.85f ′

cbwΣF = 0

Mn2 = (As −Asf )fy(d− a2 ) ΣM = 0

Total moment:

Mn =Mn1 +Mn2

25.1.2 Design, (balanced)

7 Let us derive an expression for ρb and use it for design

cd = εu

εu+εyStrainCompatibility

Asfy = .85f ′cβ1cbw + .85f ′c(b − bw)hf︸︷︷︸Asffy

ΣF = 0

thus,

Asfy = .85f ′cβ1cbw + Asffyρw = As

bwd

ρf = Asfbwd

ρw =

ρb︷︸︸︷.85f ′cfyβ1

εuεu + εy

+ρf

Hence,ρwb = ρb + ρf (25.1)ρmax = .75(ρb + ρf ) (25.2)

Example 25-1: T Beam; Moment Capacity I

For the following beam: As= 8 # 11 ( 12.48 in2); f ′c=3,000 psi; fy = 50,000 psi. Determine Mn



30"

7"

36"

14"

Solution:

1. Check requirements for isolated T sections

(a) be = 30 in should not exceed 4bw = 4(14) = 56 in√

(b) hf ≥ bu2 ⇒ 7 ≥ 14

2

√

2. Assume Rectangular section

a =(12.48)(50)

(0.85)(3)(30)= 8.16 in > hf

3. For a T sectionAsf = .85f ′

chf (b−bw)fy

= (.85)(3)(7)(30−14)50 = 5.71 in2

ρf = 5.71(14)(36) = .0113

Asw = 12.48− 5.71 = 6.77 in2

ρw = 12.48(14)(36) = .025

ρb = .85β1f ′c

fy87

87+fy

= (.85)(.85) 350

8787+50 = .0275

4. Maximum permissible ratio

ρmax = .75(ρb + ρf )= .75(.0275 + .0113) = .029 > ρw

√

5. The design moment is then obtained from

Mn1 = (5.71)(50)(36− 72 ) = 9, 280 k.in

a = (6.77)(50)(.85)(3)(14) = 9.48 in

Mn2 = (6.77)(50)(36− 9.482 ) = 10, 580 k.in

Md = (.9)(9, 280 + 10, 580) = 17, 890 k.in→ 17, 900 k.in

Example 25-2: T Beam; Moment Capacity II


Draft25.1 T Beams, (ACI 8.10) 25–5

Determine the moment capacity of the following section, assume flange dimensions to satisfy ACIrequirements; As = 6#10 = 7.59 in2; f ′c = 3 ksi; fy=60 ksi.

28"

6"

26"

10"As = 6#10= 7.59in2

Solution:

Assume rectangular beam

ρ = 7.59(28)(26) = .0104

ρb = (.85)(.85)(360

) (87

87+60

)= .0214 > ρ⇒ fs = fy

a = (7.59)(60)(.85)(3)(28) = 6.37 in > 6 in⇒ T beam

Asf = (.85)(3)(18)(6)60 = 4.59 in2

Asw = 7.59− 4.59 = 3.00 in2

ρw = 7.59(26)(10) = .0292

ρf = 4.59(26)(10) = .0177

ρmax = .75(.0214 + .0177) = .0294 > .0292⇒ Ductile failureMn1 = (4.59)(60)(26− 3) = 6, 330 k.in

As −Asf = 7.59− 4.59 = 3. in2

a = (3)(60)(.85)(3)(10) = 7.07 in

Mn2 = (3.00)(60)(26− 7.072 ) = 4, 050 k.in

Md = (.9)(6, 330 + 4, 050) = 9, 350 k.in

Example 25-3: T Beam; Design

given L = 24 ft; fy= 60 ksi; f ′c = 3 ksi; Md = 6, 400 k.in; Design a R/C T beam.



20"

3"

11"

47"

Solution:

1. Determine effective flange width:

12 (b− bw) ≤ 8hf16hf + bw = (16)(3) + 11 = 59 inL4 = 24

4 12 = 72 in

Center Line spacing = 47 in

b = 47 in

2. Assume a = 3 inAs = Md

φfy(d−a2 )

= 6,4000.9)(60)(20− 3

2 )= 6.40 in2

a = Asfy(.85)f ′

cb= (6.4)(60)

(.85)(3)(47) = 3.20 in > hf

3. Thus a T beam analysis is required.

Asf = .85f ′c(b−bw)hffy

= (.85)(3)(47−11)(3)60 = 4.58 in2

Md1 = φAsffy(d− hf2 ) = (.90)(4.58)(60)(20− 3

2 ) = 4, 570 k.in

Md2 = Md −Md1 = 6, 400− 4, 570 = 1, 830 k.in

4. Now, this is similar to the design of a rectangular section. Assume a = d5 = 20

5 = 4. in

As −Asf =1, 830

(.90)(60)(20− 4

2

) = 1.88 in2

5. checka = 1.88)(60)

(.85)(3)(11) = 4.02 in ≈ 4.00

As = 4.58 + 1.88 = 6.46 in2

ρw = 6.46(11)(20) = .0294

ρf = 4.58(11)(20) = .0208

ρb = (.85)(.85)(360

) (87

87+60

)= .0214

ρmax = .75(.0214 + .0208) = .0316 > ρw√

6. Note that 6.46 in2 (T beam) is close to As = 6.40 in2 if rectangular section was assumed.

25.2 Doubly Reinforced Rectangular Beams

8 Negative steel reinforcement is needed to

1. Increase internal moment resistance capacity (not very efficient)


Draft25.2 Doubly Reinforced Rectangular Beams 25–7

2. Support stirups

3. Reverse moments (moving load)

4. Provide ductility (earthquake)

5. Reduce creep (long term deflections)

9 Approach will again be based on a strain compatibility analysis & equilibrium equation, Fig. 25.5.

d

d’A’s

As

c

ε u .85f’c

ε’sεs

A’sf’s A’sf’y

A’sfyAsfs

.85f’c

ad-d’

(As-A’s)fs

= +

b

Figure 25.5: Doubly Reinforced Beams; Strain and Stress Diagrams

10 If ρ ≤ ρmax = .75ρb we can disregard compression steel

11 As for T beams, we decompose the tension steel into two components

1. A′s to resist the force in the top steel (assuming both yield)

2. As −A′s to resist compression in the concrete.

and we define

ρ′ =A′s

bd(25.3)

25.2.1 Tests for fs and f ′s

12 Different possibilities: Fig. 25.6

As yield?yes no

yes no yes noA’s yield? A’s yield?

I II III IV f s=fy fs=fy fs< fy fs< fy

f’s=fyf’s=fy f’s< fy f’s< fy

2

1

3

not allowed by code

{Figure 25.6: Different Possibilities for Doubly Reinforced Concrete Beams



Test 1 Is As yielding?

Assuming εs = εy, and f ′s �= fy, we have from the strain diagram

ε′s = εu − d′d (εu + εy)

f ′s = Esε′s

c = εuεu+εy

d

From equilibrium:ρbdfy = ρ′bdf ′s + .85f ′cβ1bc

Combining:

ρb = ρ1 = ρ′f ′sfy

+ .85f ′cfyβ1

εuεu + εy︸︷︷︸

ρb

thus

ρb = ρ1 = ρ′f ′sfy

+ ρb (25.4)

ρmax = 0.75ρb + ρ′f ′sfy

(25.5)Note that 0.75 premultiplies only one term as in the other

failure is ipso facto by yielding. We also note the similarity with ρmax of T Beams (where 0.75premultiplied both terms).

Test 2 fs = fy is f ′s = fy?

We set ε′s = εy, and from the strain diagram

εu

ε’s=εy

εs>εy

Figure 25.7: Strain Diagram, Doubly Reinforced Beam; is A′s Yielding?

c =εu

εu − εy d′

from equilibriumρbdfy = ρ′bdfy + .85f ′cβ1cb

combining

ρmin ≡≡ ρ2 = ρ′ + .85β1f ′cfy

d′

d

8787− fy

(25.6)

which corresponds to the minimum amount of steel to ensure yielding of compression steel atfailure. Thus, if ρ < ρmin then f ′s < fy.

Test 3 fs < fy, is f ′s = fy?

From strain diagram:c = εu

εu−εy d′

εs = d−cc−d′ εy



From equilibriumρbdfs = ρ′bdfy + .85f ′cβ1bc

combining

ρ = ρ3 =c− d′d− c

[ρ′ + .85β1

f ′cfy

c

d

](25.7)

Summary of the tests are shown in Fig. 25.8

f’s f’sf’s

fs

fsfy fyfy

fy

fy

ρρcy ρρ

< <<

==

II I III IV

1 2 3

Figure 25.8: Summary of Conditions for top and Bottom Steel Yielding

25.2.2 Moment Equations

Case I fs = fy and f ′s = fy, (small bottom and top reinforcement ratios)

Asfy = A′sfy + .85f ′cab

a = (As−A′s)fy

.85f ′cb

M In = .85f ′cab

(d− a

2

)+A′

sfy(d− d′) (25.8)

Case II We have fs = fy and f ′s < fy (small bottom and large top reinforcement ratios, most commoncase)

ε′s = εuc−d′c

f ′s = Esε′s

Asfy = A′sf

′s + .85f ′cbβ1c

solve for c and f ′s by iteration.

Using a = β1c

M IIn = .85f ′cab

(d− a

2

)+A′

sf′s(d− d′) (25.9)

Case III fs < fy and f ′s = fy (large bottom and small top reinforcement ratios, rare)

εs = εud−cc

fs = EsεsAsfs = A′

sfy + .85f ′caba = β1c



solve for a

M IIIn = .85f ′cab

(d− a

2

)+A′

sfy(d− d′) (25.10)

Case IV fs < fy and f ′s < fy (large bottom and top reinforcement ratios, rare)

ε′s = εuc−d′c

εs = εud−cc

Asfs = A′sf

′s + .85f ′cab

a = β1c

solve for a

M IVn = .85f ′cab

(d− a

2

)+A′

sfs(d− d′) (25.11)

13 Note that in most beams of normal size and proportions, it will be found that f ′s < fy when fs = fy.We nevertheless use A′

s in order to ensure ductility, stiffness and support for the stirrups.

Example 25-4: Doubly Reinforced Concrete beam; Review

Given, f ′c = 4, 000 psi, fy= 60,000 psi, A′s = 3 (1.56) = 4.68 in2, As= 4 (1.56) = 6.24 in2, determine

the moment carrying capacity of the following beam.

Solution:

ρb = (.85)β1f ′c

fy87

87+fy= (.85)(.85) 4

6087

87+60 = .0285ρ = 6.24

(16)(27.3) = .0143ρ′ = 4.68

(16)(27.3) = .0107

We can not check yet if ρ < ρmax because f ′s is yet unknown

ρmin = ρ′ + .85 f′c

fyβ1

d′d

εuεu−εy

= .0107 + (.85) 460 (.85)

327.3

.003.003− 60

29,000= .0278 > ρ

Henceρ < ρmin < ρb

.0143 < .0278 < .0285



and thus fs = fy and f ′s < fy and we have case II

solve for cf ′s = Esεu

c−d′c = (29, 000)(.003) c−3c

= 87 c−3cAsfy = A′

sf′s + .85f ′cbβ1c

(6.24)(60) = (4.68)f ′s + (.85)(4)(16)(.85)c374.4 = 4.68f ′s + 46.24cf ′s = −9.9c+ 80.2

Note that if we were to plott those two equations,

2 3 4 5 6

-100

-75

-50

-25

25

50

We note that f ′s increases with c from the strain diagram, but f ′s decreases with c from equilibrium.Graphically the solution is around 4.9.

Combining those two equations1

c2 + .7085c− 26.42 = 0

we obtain c = 4.80 in a = 0.85(4.8) = 4.078 in, and f ′s = (.003)(29, 000)4.80−34.80 = 32.6 ksi

Substituting into the moment equation

Mn = .85f ′cab(d− a

2

)+ A′

sf′s(d− d′)

= (.85)(4)(4.078)(16)(27.3− 4.078

2

)+ (4.68)(32.62)(27.3− 3)

= 9, 313 k.in

Md = 0.9(9, 313) = 8, 382 k.in = 699 k.ft

Checkρmax = .75ρb + f ′

s

fyρ′

= (.75)(.0285) + 32.660 (.0107) = .027︸︷︷︸

ρ

√

Example 25-5: Doubly Reinforced Concrete beam; Design

Given Md = 505 k.ft, f ′c = 4 ksi, fy = 60 ksi, b = 12 in, h = 24.5 in, d = 21 in, and d′ = 2.5 in,determine the reinforcement As and possibly A′

s.Solution:

Md = (505)(12) = 6, 060 k.in

ρb = .85β1f ′c

fy87

87+fy= (.85)(.85) 4

6087

87+60 = .0285ρmax = .75ρb = (.75)(.0285) = .0213Amaxs = (.0213)(12)(21) = 5.37 in2

a = Asfy.85f ′

cb= (5.37)(60)

(.85)(4)(12) = 7.89 in

Mmax = (0.9)Asfy(d− a

2

)= (.9)(5.37)(60)

(21− 7.89

2

)= 4, 943 k.in < 6, 060 k.in

1In this problem, unfortunately an iterative method diverges if we were to start with a = d5.



Thus compression steel is required. Assuming that f ′s = fy

Md2 = 6, 060− 4, 943 = 1, 117 k.in

A′s = Md2

φfy(d−d′) = 1,117(0.9)(60)(21−2.5) = 1.12 in2

⇒ A′s = 1.12 in2

As = 5.37 + 1.12 = 6.49 in2

check that f ′s = fy

ρ′ = 1.12(12)(21) = .00444

ρ = 6.49(12)(21) = .0257

ρmin = ρ′ + .85β1f ′c

fyd′d

εuεu−εy

= .00444 + (.85)(.85) 460

2.521.0

8787−60 = .0229 < ρ(.0257)

√

Note that if it turns out that f ′s < fy, then we will need to make an assumption on A′s (such as A′

s = As2 ,

as we will have three equations (2 of equilibrium and one of strain compatibility) and four unknowns(As, A′

s, f′s and c).


Draft

Chapter 26

DIRECT STIFFNESS METHOD

26.1 Introduction

26.1.1 Structural Idealization

1 Prior to analysis, a structure must be idealized for a suitable mathematical representation. Since it ispractically impossible (and most often unnecessary) to model every single detail, assumptions must bemade. Hence, structural idealization is as much an art as a science. Some of the questions confrontingthe analyst include:

1. Two dimensional versus three dimensional; Should we model a single bay of a building, or theentire structure?

2. Frame or truss, can we neglect flexural stiffness?

3. Rigid or semi-rigid connections (most important in steel structures)

4. Rigid supports or elastic foundations (are the foundations over solid rock, or over clay which mayconsolidate over time)

5. Include or not secondary members (such as diagonal braces in a three dimensional analysis).

6. Include or not axial deformation (can we neglect the axial stiffness of a beam in a building?)

7. Cross sectional properties (what is the moment of inertia of a reinforced concrete beam?)

8. Neglect or not haunches (those are usually present in zones of high negative moments)

9. Linear or nonlinear analysis (linear analysis can not predict the peak or failure load, and willunderestimate the deformations).

10. Small or large deformations (In the analysis of a high rise building subjected to wind load, themoments should be amplified by the product of the axial load times the lateral deformation, P −∆effects).

11. Time dependent effects (such as creep, which is extremely important in prestressed concrete, orcable stayed concrete bridges).

12. Partial collapse or local yielding (would the failure of a single element trigger the failure of theentire structure?).

13. Load static or dynamic (when should a dynamic analysis be performed?).

14. Wind load (the lateral drift of a high rise building subjected to wind load, is often the majorlimitation to higher structures).

15. Thermal load (can induce large displacements, specially when a thermal gradient is present.).

16. Secondary stresses (caused by welding. Present in most statically indeterminate structures).

Draft26–2 DIRECT STIFFNESS METHOD

26.1.2 Structural Discretization

2 Once a structure has been idealized, it must be discretized to lend itself for a mathematical represen-tation which will be analyzed by a computer program. This discretization should uniquely define eachnode, and member.

3 The node is characterized by its nodal id (node number), coordinates, boundary conditions, and load(this one is often defined separately), Table 26.1. Note that in this case we have two nodal coordinates,

Node No. Coor. B. C.X Y X Y Z

1 0. 0. 1 1 02 5. 5. 0 0 03 20. 5. 0 0 04 25. 2.5 1 1 1

Table 26.1: Example of Nodal Definition

and three degrees of freedom (to be defined later) per node. Furthermore, a 0 and a 1 indicate unknown orknown displacement. Known displacements can be zero (restrained) or non-zero (as caused by foundationsettlement).

4 The element is characterized by the nodes which it connects, and its group number, Table 26.2.

Element From To GroupNo. Node Node Number1 1 2 12 3 2 23 3 4 2

Table 26.2: Example of Element Definition

5 Group number will then define both element type, and elastic/geometric properties. The last one isa pointer to a separate array, Table 26.3. In this example element 1 has element code 1 (such as beamelement), while element 2 has a code 2 (such as a truss element). Material group 1 would have differentelastic/geometric properties than material group 2.

Group Element MaterialNo. Type Group1 1 12 2 13 1 2

Table 26.3: Example of Group Number

6 From the analysis, we first obtain the nodal displacements, and then the element internal forces. Thoseinternal forces vary according to the element type. For a two dimensional frame, those are the axial andshear forces, and moment at each node.

7 Hence, the need to define two coordinate systems (one for the entire structure, and one for eachelement), and a sign convention become apparent.

26.1.3 Coordinate Systems

8 We should differentiate between 2 coordinate systems:


Draft26.1 Introduction 26–3

Global: to describe the structure nodal coordinates. This system can be arbitrarily selected providedit is a Right Hand Side (RHS) one, and we will associate with it upper case axis labels, X,Y, Z,Fig. 26.1 or 1,2,3 (running indeces within a computer program).

Figure 26.1: Global Coordinate System

Local: system is associated with each element and is used to describe the element internal forces. Wewill associate with it lower case axis labels, x, y, z (or 1,2,3), Fig. 26.2.

9 The x-axis is assumed to be along the member, and the direction is chosen such that it points fromthe 1st node to the 2nd node, Fig. 26.2.

10 Two dimensional structures will be defined in the X-Y plane.

26.1.4 Sign Convention

11 The sign convention in structural analysis is completely different than the one previously adopted instructural analysis/design, Fig. 26.3 (where we focused mostly on flexure and defined a positive momentas one causing “tension below”. This would be awkward to program!).

12 In matrix structural analysis the sign convention adopted is consistent with the prevailing coordinatesystem. Hence, we define a positive moment as one which is counter-clockwise, Fig. 26.3

13 Fig. 26.4 illustrates the sign convention associated with each type of element.

14 Fig. 26.4 also shows the geometric (upper left) and elastic material (upper right) properties associatedwith each type of element.

26.1.5 Degrees of Freedom

15 A degree of freedom (d.o.f.) is an independent generalized nodal displacement of a node.

16 The displacements must be linearly independent and thus not related to each other. For example, aroller support on an inclined plane would have three displacements (rotation θ, and two translations u



Figure 26.2: Local Coordinate Systems

Figure 26.3: Sign Convention, Design and Analysis

Figure 26.4: Total Degrees of Freedom for various Type of Elements


Draft26.2 Stiffness Matrices 26–5

and v), however since the two displacements are kinematically constrained, we only have two independentdisplacements, Fig. 26.5.

Figure 26.5: Dependent Displacements

17 We note that we have been referring to generalized displacements, because we want this term toinclude translations as well as rotations. Depending on the type of structure, there may be none, oneor more than one such displacement. It is unfortunate that in most introductory courses in structuralanalysis, too much emphasis has been placed on two dimensional structures, and not enough on eitherthree dimensional ones, or two dimensional ones with torsion.

18 In most cases, there is the same number of d.o.f in local coordinates as in the global coordinate system.One notable exception is the truss element. In local coordinate we can only have one axial deformation,whereas in global coordinates there are two or three translations in 2D and 3D respectively for eachnode.

19 Hence, it is essential that we understand the degrees of freedom which can be associated with thevarious types of structures made up of one dimensional rod elements, Table 26.4.

20 This table shows the degree of freedoms and the corresponding generalized forces.

21 We should distinguish between local and global d.o.f.’s. The numbering scheme follows the followingsimple rules:

Local: d.o.f. for a given element: Start with the first node, number the local d.o.f. in the same orderas the subscripts of the relevant local coordinate system, and repeat for the second node.

Global: d.o.f. for the entire structure: Starting with the 1st node, number all the unrestrained globald.o.f.’s, and then move to the next one until all global d.o.f have been numbered, Fig. 26.6.

26.2 Stiffness Matrices

26.2.1 Truss Element

22 From strength of materials, the force/displacement relation in axial members is

σ = Eε

Aσ︸︷︷︸P

=AE

L∆︸︷︷︸1

(26.1)

Hence, for a unit displacement, the applied force should be equal to AEL . From statics, the force at the

other end must be equal and opposite.

23 The truss element (whether in 2D or 3D) has only one degree of freedom associated with each node.



Figure 26.6: Examples of Active Global Degrees of Freedom


Draft26.2 Stiffness Matrices 26–7

Type Node 1 Node 2 [k] [K](Local) (Global)

1 Dimensional{p} Fy1, Mz2 Fy3, Mz4

Beam 4× 4 4× 4{δ} v1, θ2 v3, θ4

2 Dimensional{p} Fx1 Fx2

Truss 2× 2 4× 4{δ} u1 u2{p} Fx1, Fy2, Mz3 Fx4, Fy5, Mz6

Frame 6× 6 6× 6{δ} u1, v2, θ3 u4, v5, θ6{p} Tx1, Fy2, Mz3 Tx4, Fy5, Mz6

Grid 6× 6 6× 6{δ} θ1, v2, θ3 θ4, v5, θ6

3 Dimensional{p} Fx1, Fx2

Truss 2× 2 6× 6{δ} u1, u2{p} Fx1, Fy2, Fy3, Fx7, Fy8, Fy9,

Tx4 My5, Mz6 Tx10 My11, Mz12

Frame 12× 12 12× 12{δ} u1, v2, w3, u7, v8, w9,

θ4, θ5 θ6 θ10, θ11 θ12

Table 26.4: Degrees of Freedom of Different Structure Types Systems

Hence, from Eq. 26.1, we have

[kt] =AE

L

u1 u2p1 1 −1p2 −1 1

(26.2)

26.2.2 Beam Element

24 Using Equations 23.10, 23.11, 23.13 and 23.14 we can determine the forces associated with each unitdisplacement.

[kb] =

v1 θ1 v2 θ2

V1 Eq. 23.13(v1 = 1) Eq. 23.13(θ1 = 1) Eq. 23.13(v2 = 1) Eq. 23.13(θ2 = 1)M1 Eq. 23.10(v1 = 1) Eq. 23.10(θ1 = 1) Eq. 23.10(v2 = 1) Eq. 23.10(θ2 = 1)V2 Eq. 23.14(v1 = 1) Eq. 23.14(θ1 = 1) Eq. 23.14(v2 = 1) Eq. 23.14(θ2 = 1)M2 Eq. 23.11(v1 = 1) Eq. 23.11(θ1 = 1) Eq. 23.11(v2 = 1) Eq. 23.11(θ2 = 1)

(26.3)

25 The stiffness matrix of the beam element (neglecting shear and axial deformation) will thus be

[kb] =

v1 θ1 v2 θ2V1

12EIzL3

6EIzL2 − 12EIz

L36EIzL2

M16EIzL2

4EIzL − 6EIz

L22EIzL


L212EIzL3 − 6EIz

L2

M26EIzL2

2EIzL − 6EIz

L24EIzL

(26.4)



26 We note that this is identical to Eq.23.16

V1M1

V2M2

=

v1 θ1 v2 θ2

V112EIzL3

6EIzL2 − 12EIz

L36EIzL2

M16EIzL2

4EIzL − 6EIz

L22EIzL


L212EIzL3 − 6EIz

L2

M26EIzL2

2EIzL − 6EIz

L24EIzL

︸︷︷︸k(e)

v1θ1v2θ2

(26.5)

26.2.3 2D Frame Element

27 The stiffness matrix of the two dimensional frame element is composed of terms from the truss andbeam elements where kb and kt refer to the beam and truss element stiffness matrices respectively.

[k2dfr] =

u1 v1 θ1 u2 v2 θ2

P1 kt11 0 0 kt12 0 0V1 0 kb11 kb12 0 kb13 kb14M1 0 kb21 kb22 0 kb23 kb24P2 kt21 0 0 kt22 0 0V2 0 kb31 kb32 0 kb33 kb34M2 0 kb41 kb42 0 kb43 kb44

(26.6)

Thus, we have:

[k2dfr] =

u1 v1 θ1 u2 v2 θ2P1

EAL 0 0 −EA

L 0 0V1 0 12EIz

L36EIzL2 0 − 12EIz

L36EIzL2

M1 0 6EIzL2

4EIzL 0 − 6EIz

L22EIzL

P2 −EAL 0 0 EA

L 0 0V2 0 − 12EIz

L3 − 6EIzL2 0 12EIz

L3 − 6EIL2

M2 0 6EIzL2

2EIzL 0 − 6EIz

L24EIzL

(26.7)

26.2.4 Remarks on Element Stiffness Matrices

Singularity: All the derived stiffness matrices are singular, that is there is at least one row and onecolumn which is a linear combination of others. For example in the beam element, row 4 = −row1; and L times row 2 is equal to the sum of row 3 and 6. This singularity (not present in theflexibility matrix) is caused by the linear relations introduced by the equilibrium equations whichare embedded in the formulation.

Symmetry: All matrices are symmetric due to Maxwell-Betti’s reciprocal theorem, and the stiffnessflexibility relation.

28 More about the stiffness matrix properties later.

26.3 Direct Stiffness Method

26.3.1 Orthogonal Structures

29 As a “vehicle” for the introduction to the stiffness method let us consider the problem in Fig 26.7-a,and recognize that there are only two unknown displacements, or more precisely, two global d.o.f: θ1and θ2.

30 If we were to analyse this problem by the force (or flexibility) method, then


Draft26.3 Direct Stiffness Method 26–9

Figure 26.7: Problem with 2 Global d.o.f. θ1 and θ2



1. We make the structure statically determinate by removing arbitrarily two reactions (as long as thestructure remains stable), and the beam is now statically determinate.

2. Assuming that we remove the two roller supports, then we determine the corresponding deflectionsdue to the actual laod (∆B and ∆C).

3. Apply a unit load at point B, and then C, and compute the deflections fij at note i due to a unitforce at node j.

4. Write the compatibility of displacement equation[fBB fBCfCB fCC

]{RBRC

}−{

∆1

∆2

}={

00

}(26.8)

5. Invert the matrix, and solve for the reactions

31 We will analyze this simple problem by the stiffness method.

1. The first step consists in making it kinematically determinate (as opposed to statically determinatein the flexibility method). Kinematically determinate in this case simply means restraining all thed.o.f. and thus prevent joint rotation, Fig 26.7-b.

2. We then determine the fixed end actions caused by the element load, and sum them for each d.o.f.,Fig 26.7-c: ΣFEM1 and ΣFEM2.

3. In the third step, we will apply a unit displacement (rotation in this case) at each degree offreedom at a time, and in each case we shall determine the reaction forces, K11, K21, and K12, K22

respectively. Note that we use [K], rather than k since those are forces in the global coordinatesystem, Fig 26.7-d. Again note that we are focusing only on the reaction forces corresponding toa global degree of freedom. Hence, we are not attempting to determine the reaction at node A.

4. Finally, we write the equation of equilibrium at each node:

{M1

M2

}={

ΣFEM1

ΣFEM2

}+[K11 K12

K21 K22

]{θ1θ2

}(26.9)

32 Note that the FEM being on the right hand side, they are detemined as the reactions to the appliedload. Strictly speaking, it is a load which should appear on the left hand side of the equation, and arethe nodal equivalent loads to the element load (more about this later).

33 As with the element stiffness matrix, each entry in the global stiffness matrix Kij , corresponds tothe internal force along d.o.f. i due to a unit displacement (generalized) along d.o.f. j (both in globalcoordinate systems).

Example 26-1: Beam

Considering the previous problem, Fig. 26.7-a, let P1 = 2P , M = PL, P2 = P , and P3 = P , Solvefor the displacements.Solution:

1. Using the previously defined sign convention:

ΣFEM1 = −P1L8︸︷︷︸

BA

+P2L

8︸︷︷︸BC

= −2PL8

+PL

8= −PL

8(26.10)

ΣFEM2 = −PL8︸︷︷︸

CB

(26.11)



2. If it takes 4EIL (kAB44 ) to rotate AB (Eq. 26.4) and 4EI

L (kBC22 ) to rotate BC, it will take a total forceof 8EI

L to simultaneously rotate AB and BC, (Note that a rigid joint is assumed).3. Hence, K11 which is the sum of the rotational stiffnesses at global d.o.f. 1. will be equal toK11 = 8EI

L ;similarly, K21 = 2EI

L (kBC42 ) .4. If we now rotate dof 2 by a unit angle, then we will have K22 = 4EI

L (kBC22 ) and K12 = 2EIL (kBC42 ) .

5. The equilibrium relation can thus be written as:{PL0

}︸︷︷︸

M

={ −PL

8

−PL8

}︸︷︷︸

FEM

+[

8EIL

2EIL

2EIL

4EIL

]︸︷︷︸

K

{θ1θ2

}︸︷︷︸∆

(26.12)

or {PL+ PL

8

+PL8

}=[

8EIL

2EIL

2EIL

4EIL

]{θ1θ2

}(26.13)

We note that this matrix corresponds to the structure’s stiffness matrix, and not the augmented one.6. The two by two matrix is next inverted{

θ1θ2

}=[

8EIL

2EIL

2EIL

4EIL

]−1{PL+ PL

8

+PL8

}=

{17112

PL2

EI

− 5112

PL2

EI

}(26.14)

7. Next we need to determine both the reactions and the internal forces.8. Recall that for each element {p} = [k]{δ}, and in this case {p} = {P} and {δ} = {∆} for elementAB. The element stiffness matrix has been previously derived, Eq. 26.4, and in this case the global andlocal d.o.f. are the same.9. Hence, the equilibrium equation for element AB, at the element level, can be written as:

p1p2p3p4

︸︷︷︸

{p}

=

12EIL3

6EIL2 − 12EI

L36EIL2

6EIL2

4EIL − 6EI

L22EIL

− 12EIL3 − 6EI

L212EIL3 − 6EI

L26EIL2

2EIL − 6EI

L24EIL

︸︷︷︸[k]

000

17112

PL2

EI

︸︷︷︸

{δ}

+

2P2

2PL82P2

− 2PL8

︸︷︷︸

FEM

(26.15)

solving� p1 p2 p3 p4 � = � 107

56 P3156PL

556P

514PL � (26.16)

10. Similarly, for element BC:p1p2p3p4

=

12EIL3

6EIL2 − 12EI

L36EIL2

6EIL2

4EIL − 6EI

L22EIL

− 12EIL3 − 6EI

L212EIL3 − 6EI

L26EIL2

2EIL − 6EI

L24EIL

017112

PL2

EI0

− 5112

PL2

EI

+

P2PL8P2

−PL8

(26.17)

or� p1 p2 p3 p4 � = � 7

8P914PL −P

7 0 � (26.18)

11. This simple example calls for the following observations:

1. Node A has contributions from element AB only, while node B has contributions from both ABand BC.

12. We observe that pAB3 �= pBC1 eventhough they both correspond to a shear force at node B, thedifference betweeen them is equal to the reaction at B. Similarly, pAB4 �= pBC2 due to the externallyapplied moment at node B.

2. From this analysis, we can draw the complete free body diagram, Fig. 26.7-e and then the shearand moment diagrams which is what the Engineer is most interested in for design purposes.



26.3.2 Local and Global Element Stiffness Matrices ([k(e)] [K(e)])

34 In the previous section, in which we focused on orthogonal structures, the assembly of the structure’sstiffness matrix [K(e)] in terms of the element stiffness matrices was relatively straight-forward.

35 The determination of the element stiffness matrix in global coordinates, from the element stiffnessmatrix in local coordinates requires the introduction of a transformation.

36 This section will examine the 2D transformation required to obtain an element stiffness matrix inglobal coordinate system prior to assembly (as discussed in the next section).

37 Recalling that

{p} = [k(e)]{δ} (26.19){P} = [K(e)]{∆} (26.20)

38 Let us define a transformation matrix [Γ(e)] such that:

{δ} = [Γ(e)]{∆} (26.21)

{p} = [Γ(e)]{P} (26.22)

Note that we use the same matrix Γ(e) since both {δ} and {p} are vector quantities (or tensors of orderone).

39 Substituting Eqn. 26.21 and Eqn. 26.22 into Eqn. 26.19 we obtain

[Γ(e)]{P} = [k(e)][Γ(e)]{∆} (26.23)

premultiplying by [Γ(e)]−1

{P} = [Γ(e)]−1[k(e)][Γ(e)]{∆} (26.24)

40 But since the rotation matrix is orthogonal, we have [Γ(e)]−1 = [Γ(e)]T and

{P} = [Γ(e)]T [k(e)][Γ(e)]︸︷︷︸[K(e)]

{∆} (26.25)

[K(e)] = [Γ(e)]T [k(e)][Γ(e)] (26.26)

which is the general relationship between element stiffness matrix in local and global coordinates.

26.3.2.1 2D Frame

41 The vector rotation matrix is defined in terms of 9 direction cosines of 9 different angles. Howeverfor the 2D case, Fig. 26.8, we will note that four angles are interrelated (lxX , lxY , lyX , lyY ) and can allbe expressed in terms of a single one α, where α is the direction of the local x axis (along the memberfrom the first to the second node) with respect to the global X axis. The remaining 5 terms are relatedto another angle, β, which is between the Z axis and the x-y plane. This angle is zero because we selectan orthogonal right handed coordinate system. Thus, the rotation matrix can be written as:

[γ] =

lxX lxY lxZlyX lyY lyZlzX lzY lzZ

=

cosα cos(π2 − α) 0

cos(π2 + α) cosα 00 0 1

=

cosα sinα 0− sinα cosα 0

0 0 1

(26.27)

and we observe that the angles are defined from the second subscript to the first, and that counterclock-wise angles are positive.



Figure 26.8: 2D Frame Element Rotation

42 The element rotation matrix [Γ(e)] will then be given by

p1p2p3p4p5p6

=

cosα sinα 0 0 0 0− sinα cosα 0 0 0 0

0 0 1 0 0 00 0 0 cosα sinα 00 0 0 − sinα cosα 00 0 0 0 0 1

︸︷︷︸[Γ(e)

]

P1P2P3P4P5P6

(26.28)

26.3.3 Global Stiffness Matrix

43 The physical interpretation of the global stiffness matrix K is analogous to the one of the element,i.e. If all degrees of freedom are restrained, then Kij corresponds to the force along global degree offreedom i due to a unit positive displacement (or rotation) along global degree of freedom j.

44 For instance, with reference to Fig. 26.9, we have three global degrees of freedom, ∆1, ∆2, and θ3.and the global (restrained or structure’s) stiffness matrix is

K =

K11 K12 K13

K21 K22 K23

K31 K32 K33

(26.29)

and the first column corresponds to all the internal forces in the unrestrained d.o.f. when a unit dis-placement along global d.o.f. 1 is applied.

26.3.3.1 Structural Stiffness Matrix

45 The structural stiffness matrix is assembled only for those active dgrees of freedom which are active(i.e unrestrained). It is the one which will be inverted (or rather decomposed) to determine the nodaldisplacements.

26.3.3.2 Augmented Stiffness Matrix

46 The augmented stiffness matrix is expressed in terms of all the dof. However, it is partitioned into twogroups with respective subscript ‘u’ where the displacements are known (zero otherwise), and t wherethe loads are known. {

Pt√

Ru?

}=[Ktt Ktu

Kut Kuu

]{∆t?∆u√}

(26.30)



Figure 26.9: *Frame Example (correct K32 and K33)

We note that Ktt corresponds to the structural stiffness matrix.

47 The first equation enables the calculation of the unknown displacements.

∆t = K−1tt (Pt −Ktu∆u) (26.31)

48 The second equation enables the calculation of the reactions

Ru = Kut∆t +Kuu∆u (26.32)

49 For internal book-keeping purpose, since we are assembling the augmented stiffness matrix, we proceedin two stages:

1. First number all the global unrestrained degrees of freedom

2. Then number all the global restrained degrees of freedom (i.e. those with known displacements,zero or otherwise) and multiply by -11.


50 The element internal forces (axia and shear forces, and moment at each end of the member) aredetermined from

p(e)int = k(e)δ(e) (26.33)

1An alternative scheme is to separately number the restrained dof but assign a negative number. This will enable uslater on to distinguish the restrained from unrestrained dof.



at the element level where p(e)int is the six by six array of internal forces, k(e) the element stiffness matrixin local coordinate systems, and δ(e) is the vector of nodal displacements in local coordinate system.Note that this last array is obrained by first identifying the displacements in global coordinate system,and then premultiplying it by the transofrmation matrix to obtain the displacements in local coordinatesystem.

26.3.5 Boundary Conditions, [ID] Matrix

51 Because of the boundary condition restraints, the total structure number of active degrees of freedom(i.e unconstrained) will be less than the number of nodes times the number of degrees of freedom pernode.

52 To obtain the global degree of freedom for a given node, we need to define an [ID] matrix such that:

ID has dimensions l× k where l is the number of degree of freedom per node, and k is the number ofnodes).

ID matrix is initialized to zero.

1. At input stage read ID(idof,inod) of each degree of freedom for every node such that:

ID(idof, inod) ={

0 if unrestrained d.o.f.1 if restrained d.o.f. (26.34)

2. After all the node boundary conditions have been read, assign incrementally equation numbers

(a) First to all the active dof

(b) Then to the other (restrained) dof.

(c) Multiply by -1 all the passive dof.

Note that the total number of dof will be equal to the number of nodes times the number ofdof/node NEQA.

3. The largest positive global degree of freedom number will be equal to NEQ (Number Of Equations),which is the size of the square matrix which will have to be decomposed.

53 For example, for the frame shown in Fig. 26.10:

1. The input data file may contain:

Node No. [ID]T

1 0 0 02 1 1 03 0 0 04 1 0 0

2. At this stage, the [ID] matrix is equal to:

ID =

0 1 0 1

0 1 0 00 0 0 0

(26.35)

3. After we determined the equation numbers, we would have:

ID =

1 −1 5 −3

2 −2 6 83 4 7 9

(26.36)



Figure 26.10: Example for [ID] Matrix Determination

26.3.6 LM Vector

54 The LM vector of a given element gives the global degree of freedom of each one of the element degreeof freedom’s. For the structure shown in Fig. 26.10, we would have:

�LM� = � −1 −2 4 5 6 7 � element 1 (2 → 3)�LM� = � 5 6 7 1 2 3 � element 2 (3 → 1)�LM� = � 1 2 3 −3 8 9 � element 3 (1 → 4)

26.3.7 Assembly of Global Stiffness Matrix

55 As for the element stiffness matrix, the global stiffness matrix [K] is such that Kij is the force indegree of freedom i caused by a unit displacement at degree of freedom j.

56 Whereas this relationship was derived from basic analysis at the element level, at the structure level,this term can be obtained from the contribution of the element stiffness matrices [K(e)] (written in globalcoordinate system).

57 For each Kij term, we shall add the contribution of all the elements which can connect degree offreedom i to degree of freedom j, assuming that those forces are readily available from the individualelement stiffness matrices written in global coordinate system.

58 Kij is non-zero if and only if degree of freedom i and degree of freedom j are connected by an elementor share a node.

59 There are usually more than one element connected to a dof. Hence, individual element stiffnessmatrices terms must be added up.

60 Because each term of all the element stiffness matrices must find its position inside the global stiff-ness matrix [K], it is found computationally most effective to initialize the global stiffness matrix[KS ](NEQA×NEQA) to zero, and then loop through all the elements, and then through each entry ofthe respective element stiffness matrix K(e)

ij .

61 The assignment of the element stiffness matrix term K(e)ij (note that e, i, and j are all known since

we are looping on e from 1 to the number of elements, and then looping on the rows and columns of the



element stiffness matrix i, j) into the global stiffness matrix KSkl is made through the LM vector (note

that it is k and l which must be determined).

62 Since the global stiffness matrix is also symmetric, we would need to only assemble one side of it,usually the upper one.

63 Contrarily to the previous method, we will assemble the full augmented stiffness matrix.

Example 26-2: Assembly of the Global Stiffness Matrix

As an example, let us consider the frame shown in Fig. 26.11.

��

��

��

��

7.416 m 8 m

3 m

4 kN/m50kN

8 m

Figure 26.11: Simple Frame Anlysed with the MATLAB Code

The ID matrix is initially set to:

[ID] =

1 0 1

1 0 11 0 1

(26.37)

We then modify it to generate the global degrees of freedom of each node:

[ID] =

−4 1 −7−5 2 −8−6 3 −9

(26.38)

Finally the LM vectors for the two elements (assuming that Element 1 is defined from node 1 to node2, and element 2 from node 2 to node 3):

[LM ] =[ −4 −5 −6 1 2 3

1 2 3 −7 −8 −9

](26.39)

Let us simplfy the operation by designating the element stiffness matrices in global coordinates as follows:

K(1) =

−4 −5 −6 1 2 3−4 A11 A12 A13 A14 A15 A16

−5 A21 A22 A23 A24 A25 A26

−6 A31 A32 A33 A34 A35 A36

1 A41 A42 A43 A44 A45 A46

2 A51 A52 A53 A54 A55 A56

3 A61 A62 A63 A64 A65 A66

(26.40-a)

K(2) =

1 2 3 −7 −8 −91 B11 B12 B13 B14 B15 B16

2 B21 B22 B23 B24 B25 B26

3 B31 B32 B33 B34 B35 B36

−7 B41 B42 B43 B44 B45 B46

−8 B51 B52 B53 B54 B55 B56

−9 B61 B62 B63 B64 B65 B66

(26.40-b)



We note that for each element we have shown the corresponding LM vector.Now, we assemble the global stiffness matrix

K =

A44 +B11 A45 +B12 A46 +B13 A41 A42 A43 B14 B15 B16

A54 +B21 A55 +B22 A56 +B23 A51 A52 A53 B24 B25 B26

A64 +B31 A65 +B32 A66 +B33 A61 A62 A63 B34 B35 B36

A14 A15 A16 A11 A12 A13 0 0 0A24 A25 A26 A21 A22 A23 0 0 0A34 A35 A36 A31 A32 A33 0 0 0B41 B42 B43 0 0 0 B44 B45 B46

B51 B52 B53 0 0 0 B54 B55 B56

B61 B62 B63 0 0 0 B64 B65 B66

(26.41)

We note that some terms are equal to zero because we do not have a connection between the corre-sponding degrees of freedom (i.e. node 1 is not connected to node 3).

26.3.8 Algorithm

64 The direct stiffness method can be summarized as follows:

Preliminaries: First we shall

1. Identify type of structure (beam, truss, grid or frame) and determine the

(a) Number of spatial coordinates (1D, 2D, or 3D)(b) Number of degree of freedom per node (local and global)(c) Number of cross-sectional and material properties

2. Determine the global unrestrained and restrained degree of freedom equation numbers foreach node, Update the [ID] matrix (which included only 0’s and 1’s in the input data file).

Analysis :

1. For each element, determine

(a) Vector LM relating local to global degree of freedoms.(b) Element stiffness matrix [k(e)](c) Angle α between the local and global x axes.(d) Rotation matrix [Γ(e)](e) Element stiffness matrix in global coordinates [K(e)] = [Γ(e)]T [k(e)][Γ(e)]

2. Assemble the augmented stiffness matrix [K(S)] of unconstrained and constrained degree offreedom’s.

3. Extract [Ktt] from [K(S)] and invert (or decompose into into [Ktt] = [L][L]T where [L] is alower triangle matrix.

4. Assemble load vector {P} in terms of nodal load and fixed end actions.

5. Backsubstitute and obtain nodal displacements in global coordinate system.

6. SOlve for the reactions.

7. For each element, transform its nodal displacement from global to local coordinates {δ} =[Γ(e)]{∆}, and determine the internal forces [p] = [k]{δ}.

65 Some of the prescribed steps are further discussed in the next sections.

Example 26-3: Direct Stiffness Analysis of a Truss



8

7

65

4

3

2

1

54

321

100k

50k

12’

16’16’

Figure 26.12:

Using the direct stiffness method, analyse the truss shown in Fig. 26.12.Solution:

1. Determine the structure ID matrix

Node # Bound. Cond.X Y

1 0 12 0 03 1 14 0 05 0 0

ID =[

0 0 1 0 01 0 1 0 0

](26.42-a)

=[Node 1 2 3 4 5

1 2 −2 4 6−1 3 −3 5 7

](26.42-b)

2. The LM vector of each element is evaluated next

�LM�1 = � 1 −1 4 5 � (26.43-a)�LM�2 = � 1 −1 2 3 � (26.43-b)�LM�3 = � 2 3 4 5 � (26.43-c)�LM�4 = � 4 5 6 7 � (26.43-d)�LM�5 = � −2 −3 4 5 � (26.43-e)�LM�6 = � 2 3 6 7 � (26.43-f)�LM�7 = � 2 3 0 0 � (26.43-g)�LM�8 = � −2 −3 6 7 � (26.43-h)

3. Determine the element stiffness matrix of each element in the global coordinate system noting thatfor a 2D truss element we have

[K(e)] = [Γ(e)]T [k(e)][Γ(e)] (26.44-a)

=EA

L

c2 cs −c2 −cscs s2 −cs −s2−c2 −cs c2 cs−cs −s2 cs s2

(26.44-b)



where c = cosα = x2−x1L ; s = sinα = Y2−Y1

L

Element 1 L = 20′, c = 16−0

20 = 0.8, s = 12−020 = 0.6, EAL = (30,000 ksi)(10 in2)

20′ = 15, 000 k/ft.

[K1] =

1 −1 4 51 9600 7200 −9600 −7200−1 7200 5400 −7200 −54004 −9600 −7200 9600 72005 −7200 −5400 7200 5400

(26.45)

Element 2 L = 16′, c = 1 , s = 0 , EA

L = 18, 750 k/ft.

[K2] =

1 −1 2 31 18, 750 0 −18, 750 0−1 0 0 0 02 −18, 750 0 18, 750 03 0 0 0 0

(26.46)

Element 3 L = 12′, c = 0 , s = 1 , EA

L = 25, 000 k/ft.

[K3] =

2 3 4 52 0 0 0 03 0 25, 000 0 −25, 0004 0 0 0 05 0 −25, 000 0 25, 000

(26.47)

Element 4 L = 16′, c = 1 , s = 0 , EA

L = 18, 750 k/ft.

[K4] =

4 5 6 74 18, 750 0 −18, 750 05 0 0 0 06 −18, 750 0 18, 750 07 0 0 0 0

(26.48)

Element 5 L = 20′, c = −16−0

20 = − 0.8 , s = 0.6 , EAL = 15, 000 k/ft.

[K5] =

−2 −3 4 5

−2 9600 −7200 −9600 7200−3 −7200 5400 7200 −54004 −9600 7200 9600 −72005 7200 −5400 −7200 5400

(26.49)

Element 6 L = 20′, c = 0.8 , s = 0.6 , EA

L = 15, 000 k/ft.

[K6] =

2 3 6 72 9600 7200 −9600 −72003 7200 5400 −7200 −54006 −9600 −7200 9600 72007 −7200 −5400 7200 5400

(26.50)

Element 7 L = 16′, c = 1 , s = 0 , EA

L = 18, 750 k/ft.

[K7] =

2 3 0 02 18, 750 0 −18, 750 03 0 0 0 00 −18, 750 0 18, 750 00 0 0 0 0

(26.51)



Element 8 L = 12′, c = 0 , s = 1 , EA

L = 25, 000 k/ft.

[K8] =

−2 −3 6 7

−2 0 0 0 0−3 0 25, 000 0 −25, 0006 0 0 0 07 0 −25, 000 0 25, 000

(26.52)

4. Assemble the global stiffness matrix in k/ft Note that we are not assembling the augmented stiffnessmatrix, but rather its submatrix [Ktt].

00

−100k0050k0

=

9600 + 18, 750 −18, 750 0 −9600 −7200 0 09600 + (2) 18, 750 7200 0 0 −9600 −7200

5400 + 25, 000 0 −25, 000 −7200 −540018, 750 + (2)9600 7200 − 7200 −18, 750 0

SYMMETRIC 25, 000 + 5400(2) 0 018, 750 + 9600 7200

25, 000 + 5400

u1u2v3u4v5u6v7

(26.53)

5. convert to k/in and simplify

00−100

00500

=

2362.5 −1562.5 0 −800 −600 0 03925.0 600 0 0 −800 −600

2533.33 0 −2083.33 −600 −4503162.5 0 −1562.5 0

SYMMETRIC 2983.33 0 02362.5 600

2533.33

U1U2V3U4V5U6V7

(26.54)6. Invert stiffness matrix and solve for displacements

U1U2V3U4V5U6V7

=

−0.0223 in

0.00433 in

−0.116 in

−0.0102 in

−0.0856 in

−0.00919 in

−0.0174 in

(26.55)

7. Solve for member internal forces (in this case axial forces) in local coordinate systems

{u1u2

}=[

c s −c −s−c −s c s

]U1V1U2V2

(26.56)

Element 1

{p1p2

}1

= (15, 000 k/ft)(112

ft

in)[

0.8 0.6 −0.8 −0.6−0.8 −0.6 0.8 0.6

]−0.0223

0−0.0102−0.0856

(26.57-a)

={

52.1 k

−52.1 k

}Compression (26.57-b)

Element 2

{p1p2

}2

= 18, 750 k/ft(112

ft

in)[

1 0 −1 0−1 0 1 0

]−0.0233

00.00433−0.116

(26.58-a)



={ −43.2 k

43.2 k

}Tension (26.58-b)

Element 3

{p1p2

}3

= 25, 000 k/ft(112

ft

in)[

0 1 0 −10 −1 0 1

]

0.00433−0.116−0.0102−0.0856

(26.59-a)

={ −63.3 k

63.3 k

}Tension (26.59-b)

Element 4

{p1p2

}4

= 18, 750 k/ft(112

ft

in)[

1 0 −1 0−1 0 1 0

]−0.0102−0.0856−0.00919−0.0174

(26.60-a)

={ −1.58 k

1.58 k

}Tension (26.60-b)

Element 5

{p1p2

}5= 15, 000 k/ft(

112

ft

in)[ −0.8 0.6 0.8 −0.6

0.8 −0.6 −0.8 0.6

]{ −0.0102−0.0856

}(26.61-a)

={

54.0 k

−54.0 k


Element 6

{p1p2

}6= 15, 000 k/ft(

112

ft

in)[

0.8 0.6 −0.8 −0.6−0.8 −0.6 0.8 0.6

]−0.116−0.00919−0.0174

(26.62-a)

={ −60.43 k

60.43 k

}Tension (26.62-b)

Element 7

{p1p2

}7= 18, 750 k/ft(

112

ft

in)[

1 0 −1 0−1 0 1 0

]−0.116

00

(26.63-a)

={

6.72 k

−6.72 k


Element 8

{p1p2

}8

= 25, 000 k/ft(112

ft

in)[

0 1 0 −10 −1 0 1

]

00−0.00919−0.0174

(26.64-a)

={

36.3 k

−36.3 k




8. Determine the structure’s MAXA vector

[K] =

1 3 9 142 5 8 13 19 25

4 7 12 18 246 11 17 23

10 16 2215 21

20

MAXA =

1246101520

(26.65)

Thus, 25 terms would have to be stored.

Example 26-4: Analysis of a Frame with MATLAB

The simple frame shown in Fig. 26.13 is to be analysed by the direct stiffness method. Assume:E = 200, 000 MPa, A = 6, 000 mm2, and I = 200×106 mm4. The complete MATLAB solution is shownbelow along with the results.

��

��

��

��

7.416 m 8 m

3 m

4 kN/m50kN

8 m

Figure 26.13: Simple Frame Anlysed with the MATLAB Code

% zero the matricesk=zeros(6,6,2);K=zeros(6,6,2);Gamma=zeros(6,6,2);% Structural properties units: mm^2, mm^4, and MPa(10^6 N/m)A=6000;II=200*10^6;EE=200000;% Convert units to meter and kNA=A/10^6;II=II/10^12;EE=EE*1000;% Element 1i=[0,0];j=[7.416,3];[k(:,:,1),K(:,:,1),Gamma(:,:,1)]=stiff(EE,II,A,i,j);% Element 2i=j;j=[15.416,3];[k(:,:,2),K(:,:,2),Gamma(:,:,2)]=stiff(EE,II,A,i,j);% Define ID matrixID=[

-4 1 -7;-5 2 -8;-6 3 -9];

% Determine the LM matrixLM=[

-4 -5 -6 1 2 3;1 2 3 -7 -8 -9];

% Assemble augmented stiffness matrixKaug=zeros(9);



for elem=1:2for r=1:6

lr=abs(LM(elem,r));for c=1:6

lc=abs(LM(elem,c));Kaug(lr,lc)=Kaug(lr,lc)+K(r,c,elem);

endend

end% Extract the structures Stiffness MatrixKtt=Kaug(1:3,1:3);% Determine the fixed end actions in local coordinate systemfea(1:6,1)=0;fea(1:6,2)=[0,8*4/2,4*8^2/12,0,8*4/2,-4*8^2/12]’;% Determine the fixed end actions in global coordinate systemFEA(1:6,1)=Gamma(:,:,1)*fea(1:6,1);FEA(1:6,2)=Gamma(:,:,2)*fea(1:6,2);% FEA_Rest for all the restrained nodesFEA_Rest=[0,0,0,FEA(4:6,2)’];% Assemble the load vector for the unrestrained nodeP(1)=50*3/8;P(2)=-50*7.416/8-fea(2,2);P(3)=-fea(3,2);% Solve for the Displacements in meters and radiansDisplacements=inv(Ktt)*P’% Extract KutKut=Kaug(4:9,1:3);% Compute the Reactions and do not forget to add fixed end actionsReactions=Kut*Displacements+FEA_Rest’% Solve for the internal forces and do not forget to include the fixed end actionsdis_global(:,:,1)=[0,0,0,Displacements(1:3)’];dis_global(:,:,2)=[Displacements(1:3)’,0,0,0];for elem=1:2

dis_local=Gamma(:,:,elem)*dis_global(:,:,elem)’;int_forces=k(:,:,elem)*dis_local+fea(1:6,elem)

end

function [k,K,Gamma]=stiff(EE,II,A,i,j)% Determine the lengthL=sqrt((j(2)-i(2))^2+(j(1)-i(1))^2);% Compute the angle theta (carefull with vertical members!)if(j(1)-i(1))~=0

alpha=atan((j(2)-i(2))/(j(1)-i(1)));else

alpha=-pi/2;end% form rotation matrix GammaGamma=[cos(alpha) sin(alpha) 0 0 0 0;-sin(alpha) cos(alpha) 0 0 0 0;0 0 1 0 0 0;0 0 0 cos(alpha) sin(alpha) 0;0 0 0 -sin(alpha) cos(alpha) 0;0 0 0 0 0 1];% form element stiffness matrix in local coordinate systemEI=EE*II;EA=EE*A;k=[EA/L, 0, 0, -EA/L, 0, 0;

0, 12*EI/L^3, 6*EI/L^2, 0, -12*EI/L^3, 6*EI/L^2;



0, 6*EI/L^2, 4*EI/L, 0, -6*EI/L^2, 2*EI/L;-EA/L, 0, 0, EA/L, 0, 0;

0, -12*EI/L^3, -6*EI/L^2, 0, 12*EI/L^3, -6*EI/L^2;0, 6*EI/L^2, 2*EI/L, 0, -6*EI/L^2, 4*EI/L];

% Element stiffness matrix in global coordinate systemK=Gamma’*k*Gamma;

This simple proigram will produce the following results:

Displacements =

0.0010-0.0050-0.0005

Reactions =

130.497355.676613.3742

-149.247322.6734

-45.3557

int_forces = int_forces =

141.8530 149.24732.6758 9.326613.3742 -8.0315

-141.8530 -149.2473-2.6758 22.67348.0315 -45.3557

We note that the internal forces are consistent with the reactions (specially for the second node ofelement 2), and amongst themselves, i.e. the moment at node 2 is the same for both elements (8.0315).

Example 26-5: Analysis of a simple Beam with Initial Displacements

The full stiffness matrix of a beam element is given by

[K] =

v1 θ1 v2 θ2

V1 12EI/L3 6EI/L2 −12EI/L3 6EI/L2

M1 6EI/L2 4EI/L −6EI/L2 2EI/LV2 −12EI/L3 −6EI/L2 12EI/L3 −6EI/L2

M2 6EI/L2 2EI/L −6EI/L2 4EI/L

(26.66)

This matrix is singular, it has a rank 2 and order 4 (as it embodies also 2 rigid body motions).

66 We shall consider 3 different cases, Fig. 26.14

Cantilivered Beam/Point Load

1. The element stiffness matrix is

k =

−3 −4 1 2

−3 12EI/L3 6EI/L2 −12EI/L3 6EI/L2

−4 6EI/L2 4EI/L −6EI/L2 2EI/L1 −12EI/L3 −6EI/L2 12EI/L3 −6EI/L2

2 6EI/L2 2EI/L −6EI/L2 4EI/L



Figure 26.14: ID Values for Simple Beam

2. The structure stiffness matrix is assembled

K =

1 2 −3 −4

1 12EI/L2 −6EI/L2 −12EI/L3 −6EI/L2

2 −6EI/L2 4EI/L 6EI/L2 2EI/L−3 −12EI/L3 6EI/L2 12EI/L3 6EI/L2

−4 −6EI/L2 2EI/L 6EI/L2 4EI/L

3. The global matrix can be rewritten as

−P√0√R3?R4?

=

12EI/L2 −6EI/L2 −12EI/L3 −6EI/L2

−6EI/L2 4EI/L 6EI/L2 2EI/L−12EI/L3 6EI/L2 12EI/L3 6EI/L2

−6EI/L2 2EI/L 6EI/L2 4EI/L

∆1?θ2?∆3

√θ4

√

4. Ktt is inverted (or actually decomposed) and stored in the same global matrix

L3/3EI L2/2EI −12EI/L3 −6EI/L2

L2/2EI L/EI 6EI/L2 2EI/L

−12EI/L3 6EI/L2 12EI/L3 6EI/L2


5. Next we compute the equivalent load, P′t = Pt −Ktu∆u, and overwrite Pt by P′

t

Pt −Ktu∆u =

−P0

00

−





−P0

0

0

=

−P0

00

6. Now we solve for the displacement ∆t = K−1tt P

′t, and overwrite Pt by ∆t

∆1

θ2

00

=





−P0

00

=

−PL3/3EI

−PL2/2EI

00



7. Finally, we solve for the reactions, Ru = Kut∆tt +Kuu∆u, and overwrite ∆u by Ru

−PL3/3EI−PL2/2EI

R3

R4

=





−PL3/3EI

−PL2/2EI

0

0

=

−PL3/3EI−PL2/2EI

P

PL

Simply Supported Beam/End Moment


k =

−3 1 −4 2

−3 12EI/L3 6EI/L2 −12EI/L3 6EI/L2

1 6EI/L2 4EI/L −6EI/L2 2EI/L−4 −12EI/L3 −6EI/L2 12EI/L3 −6EI/L2



K =

1 2 −3 −4

1 4EI/L 2EI/L 6EI/L2 −6EI/L2

2 2EI/L 4EI/L 6EI/L2 −6EI/L2

−3 6EI/L2 6EI/L2 12EI/L3 −12EI/L3

−4 −6EI/L2 −6EI/L2 −12EI/L3 12EI/L3

3. The global stiffness matrix can be rewritten as

0√

M√

R3?R4?

=

4EI/L 2EI/L 6EI/L2 −6EI/L2

2EI/L 4EI/L 6EI/L2 −6EI/L2

6EI/L2 6EI/L2 12EI/L3 −12EI/L3

−6EI/L2 −6EI/L2 −12EI/L3 12EI/L3

θ1?θ2?∆3

√∆4

√

4. Ktt is inverted

L3/3EI −L/6EI 6EI/L2 −6EI/L2

−L/6EI L/3EI 6EI/L2 −6EI/L2


−6EI/L2 −6EI/L2 −12EI/L3 12EI/L3

5. We compute the equivalent load, P′t = Pt −Ktu∆u, and overwrite Pt by P′

t

Pt −Ktu∆u =

0

M

00

−




−6EI/L2 −6EI/L2 −12EI/L3 12EI/L3

0M

0

0

=

0

M

00

6. Solve for the displacements, ∆t = K−1tt P


θ1

θ2

00

=




−6EI/L2 −6EI/L2 −12EI/L3 12EI/L3

0

M

00

=

−ML/6EIML/3EI

00



7. Solve for the reactions, Rt = Kut∆tt +Kuu∆u, and overwrite ∆u by Ru

−ML/6EIML/3EI

R1

R2

=




−6EI/L2 −6EI/L2 −12EI/L3 12EI/L3

−ML/6EIML/3EI

0

0

=

−ML/6EIML/3EI

M/L

−M/L

Cantilivered Beam/Initial Displacement and Concentrated Moment


k =

−2 −3 −4 1

−2 12EI/L3 6EI/L2 −12EI/L3 6EI/L2

−3 6EI/L2 4EI/L −6EI/L2 2EI/L−4 −12EI/L3 −6EI/L2 12EI/L3 −6EI/L2



K =

1 −2 −3 −4

1 4EI/L 6EI/L2 2EI/L −6EI/L2

−2 6EI/L2 12EI/L3 6EI/L2 −12EI/L3

−3 2EI/L 6EI/L2 4EI/L −6EI/L2

−4 −6EI/L2 −12EI/L3 −6EI/L2 12EI/L3

3. The global matrix can be rewritten as

M√

R2?R3?R4?

=

4EI/L 6EI/L2 2EI/L −6EI/L2



−6EI/L2 −12EI/L3 −6EI/L2 12EI/L3

θ1?∆2

√θ3

√∆4

√

4. Ktt is inverted (or actually decomposed) and stored in the same global matrix L/4EI 6EI/L2 2EI/L −6EI/L2



−6EI/L2 −12EI/L3 −6EI/L2 12EI/L3

5. Next we compute the equivalent load, P′t = Pt −Ktu∆u, and overwrite Pt by P′

t

Pt −Ktu∆u =

M

00∆0

−

L/4EI 6EI/L2 2EI/L −6EI/L2



−6EI/L2 −12EI/L3 −6EI/L2 12EI/L3

θ1

0

0

∆0

=

M + 6EI∆0/L2

00∆0

6. Now we solve for the displacements, ∆t = K−1tt P


θ1

00∆0

=




−6EI/L2 −12EI/L3 −6EI/L2 12EI/L3

M + 6EI∆0/L2

00∆0

=

ML/4EI + 3∆0/2L

00∆0


Draft26.4 Computer Program Organization 26–29

7. Finally, we solve for the reactions, Rt = Kut∆tt +Kuu∆u, and overwrite ∆u by Ru

ML/4EI + 3∆0/2L

R2

R3

R4

=




−6EI/L2 −12EI/L3 −6EI/L2 12EI/L3

ML/4EI + 3∆0/2L

0

0

∆0

=

ML/4EI + 3∆0/2L

3M/2L− 3EI∆0/L3

M/2 − 3EI∆0/L2

−3M/2L + 3EI∆0/L3

26.4 Computer Program Organization

66 The main program should,

1. Read

(a) title card

(b) control card which should include:

i. Number of nodesii. Number of elementsiii. Type of structure: beam, grid, truss, or frame; (2D or 3D)iv. Number of different element propertiesv. Number of load cases

2. Determine:

(a) Number of spatial coordinates for the structure

(b) Number of local and global degrees of freedom per node

3. For each node read:

(a) Node number

(b) Boundary conditions of each global degree of freedom [ID]

(c) Spatial coordinates

Note that all the above are usually written on the same “data card”

4. For each element, read:

(a) Element number

(b) First and second node

(c) Element Property number

5. For each element property group read the associated elastic and cross sectional characteristics.Note these variables will depend on the structure type.

6. Determine the vector ∆u which stores the initial displacements.

7. Loop over all the elements and for each one:



(a) Retrieve its properties

(b) Determine the length

(c) Call the appropriate subroutines which will determine:

i. The stiffness matrix in local coordinate systems [k(e)].ii. The angle α and the transformation matrix [Γ(e)].

8. Assembly of the global stiffness matrix

(a) Initialize the global stiffness matrix to zero

(b) Loop through each element, e, and for each element:

i. Retrieve its stiffness matrix (in local coordinates) [k(e)] and transformation matrix [Γ(e)].ii. Compute the element stiffness matrix in global coordinates from [K(e)] = [Γ(e)]T [k(e)][Γ(e)].iii. Define the {LM} array of the elementiv. Loop through each row i and column j of the element stiffness matrix, and for those

degree of freedom not equal to zero, add the contributions of the element to the structure’sstiffness matrix KS [LM(i), LM(j)] = KS [LM(i), LM(j)] +K(e)[i, j]

9. Extract the structure’s stiffness matrix [Ktt] from the augmented stiffness matrix.

10. Invert the structure’s stiffness matrix (or decompose it).

11. For each load case:

(a) Determine the nodal equivalent loads (fixed end actions), if any.

(b) Assemble the load vector

(c) Load assembly (once for each load cae) once the stiffness matrix has been decomposed, thanthe main program should loop through each load case and,

i. Initialize the load vector (of length NEQ) to zero.ii. Read number of loaded nodes. For each loaded node store the non-zero values inside the

load vector (using the [ID] matrix for determining storage location).iii. Loop on all loaded elements:

A. Read element number, and load valueB. Compute the fixed end actions and rotate them from local to global coordinates.C. Using the LM vector, add the fixed end actions to the nodal load vector (unless the

corresponding equation number is zero, ie. restrained degree of freedom).D. Store the fixed end actions for future use.

(d) Apply Eq. 26.31 to determine the nodal displacements ∆t = K−1tt (Pt −Ktu∆u)

(e) Apply Eq. 26.32 to determine the nodal reactions Rt = Kut∆t +Kuu∆u

(f) Determine the internal forces (axial, shear and moment)

i. For each element retrieve:A. nodal coordinatesB. rotation matrix [Γ(e)].C. element stiffness matrix [k(e)].

ii. Compute nodal displacements in local coordinate system from{

δ(e)}

= [Γ(e)] {∆}iii. Compute element internal forces from {p} = [k(e)]

{δ(e)

}iv. If the element is loaded, add corresponding fixed end actionsv. print the interior forces


Draft26.5 Computer Implementation with MATLAB 26–31

26.5 Computer Implementation with MATLAB

67 You will be required, as part of your term project, to write a simple MATLAB (or whatever otherlanguage you choose) program for the analysis of two dimensional frames with nodal load and initialdisplacement, as well as element load.

68 To facilitate the task, your instructor has taken the liberty of taking a program written by Mr. DeanFrank (as part of his term project with this instructor in the Advanced Structural Analysis course, Fall1995), modified it with the aid of Mr. Pawel Smolarki, and is making available most, but not all of it toyou. Hence, you will be expected to first familiarize yourself with the code made available to you, andthen complete it by essentially filling up the missing parts.

26.5.1 Program Input

From Dean Frank’s User’s Manual

69 It is essential that the structure be idealized such that it can be discretized. This discretization shoulddefine each node and element uniquely. In order to decrease the required amount of computer storageand computation it is best to number the nodes in a manner that minimizes the numerical separation ofthe node numbers on each element. For instance, an element connecting nodes 1 and 4, could be betterdefined by nodes 1 and 2, and so on. As it was noted previously, the user is required to have a decentunderstanding of structural analysis and structural mechanics. As such, it will be necessary for the userto generate or modify an input file input.m using the following directions. Open the file called input.mand set the existing variables in the file to the appropriate values. The input file has additional helpfuldirections given as comments for each variable. After setting the variables to the correct values, be sureto save the file. Please note that the program is case-sensitive.

70 In order for the program to be run, the user must supply the required data by setting certain variablesin the file called indat.m equal to the appropriate values. All the user has to do is open the text filecalled indat.txt, fill in the required values and save the file as indat.m in a directory within MATAB’spath. There are helpful hints within this file. It is especially important that the user keep track ofunits for all of the variables in the input data file. All of the units MUST be consistent. It is suggestedthat one always use the same units for all problems. For example, always use kips and inches, or kilo-newtons and millimeters.

26.5.1.1 Input Variable Descriptions

71 A brief description of each of the variables to be used in the input file is given below:npoin This variable should be set equal to the number of nodes that comprise the structure. A node

is defined as any point where two or more elements are joined.nelem This variable should be set equal to the number of elements in the structure. Elements are

the members which span between nodes.istrtp This variable should be set equal to the type of structure. There are six types of structures

which this program will analyze: beams, 2-D trusses, 2-D frames, grids, 3-D trusses, and 3-D frames. Setthis to 1 for beams, 2 for 2D-trusses, 3 for 2D- frames, 4 for grids, 5 for 3D-trusses, and 6 for 3D-frames.An error will occur if it is not set to a number between 1 and 6. Note only istrp=3 was kept.

nload This variable should be set equal to the number of different load cases to be analyzed. A loadcase is a specific manner in which the structure is loaded.

ID (matrix) The ID matrix contains information concerning the boundary conditions for each node.The number of rows in the matrix correspond with the number of nodes in the structure and the numberof columns corresponds with the number of degrees of freedom for each node for that type of structuretype. The matrix is composed of ones and zeros. A one indicates that the degree of freedom is restrainedand a zero means it is unrestrained.

nodecoor (matrix) This matrix contains the coordinates (in the global coordinate system) of thenodes in the structure. The rows correspond with the node number and the columns correspond withthe global coordinates x, y, and z, respectively. It is important to always include all three coordinatesfor each node even if the structure is only two- dimensional. In the case of a two-dimensional structure,the z-coordinate would be equal to zero.



lnods (matrix) This matrix contains the nodal connectivity information. The rows correspond withthe element number and the columns correspond with the node numbers which the element is connectedfrom and to, respectively.

E,A,Iy (arrays) These are the material and cross-sectional properties for the elements. They arearrays with the number of terms equal to the number of elements in the structure. The index number ofeach term corresponds with the element number. For example, the value of A(3) is the area of element3, and so on. E is the modulus of elasticity, A is the cross-sectional area, Iy is the moment of inertiaabout the y axes

Pnods This is an array of nodal loads in global degrees of freedom. Only put in the loads in theglobal degrees of freedom and if there is no load in a particular degree of freedom, then put a zero in itsplace. The index number corresponds with the global degree of freedom.

Pelem This an array of element loads, or loads which are applied between nodes. Only one loadbetween elements can be analyzed. If there are more than one element loads on the structure, theequivalent nodal load can be added to the nodal loads. The index number corresponds with the elementnumber. If there is not a load on a particular member, put a zero in its place. These should be in localcoordinates.

a This is an array of distances from the left end of an element to the element load. The index numbercorresponds to the element number. If there is not a load on a particular member, put a zero in itsplace. This should be in local coordinates.

w This is an array of distributed loads on the structure. The index number corresponds with theelement number. If there is not a load on a particular member, put a zero in its place. This should bein local coordinates

dispflag Set this variable to 1 if there are initial displacements and 0 if there are none.initial displ This is an array of initial displacements in all structural degrees of freedom. This

means that you must enter in values for all structure degrees of freedom, not just those restrained. Forexample, if the structure is a 2D truss with 3 members and 3 node, there would be 6 structural degreesof freedom, etc. If there are no initial displacements, then set the values equal to zero.

angle This is an array of angles which the x-axis has possibly been rotated. This angle is takenas positive if the element has been rotated towards the z-axis. The index number corresponds to theelement number.

drawflag Set this variable equal to 1 if you want the program to draw the structure and 0 if you donot.

26.5.1.2 Sample Input Data File

The contents of the input.m file which the user is to fill out is given below:

%**********************************************************************************************

% Scriptfile name: indat.m (EXAMPLE 2D-FRAME INPUT DATA)

%

% Main Program: casap.m

%

% This is the main data input file for the computer aided

% structural analysis program CASAP. The user must supply

% the required numeric values for the variables found in

% this file (see user’s manual for instructions).

%

% By Dean A. Frank

% CVEN 5525 - Term Project

% Fall 1995

%

% Edited by Pawel Smolarkiewicz, 3/16/99

% Simplified for 2D Frame Case only

%

%**********************************************************************************************

% HELPFUL INSTRUCTION COMMENTS IN ALL CAPITALS

% SET NPOIN EQUAL TO THE NUMBER OF NODES IN THE STRUCTURE

npoin=3;

% SET NELEM EQUAL TO THE NUMBER OF ELEMENTS IN THE STRUCTURE



nelem=2;

% SET NLOAD EQUAL TO THE NUMBER OF LOAD CASES

nload=1;

% INPUT THE ID MATRIX CONTAINING THE NODAL BOUNDARY CONDITIONS (ROW # = NODE #)

ID=[1 1 1;

0 0 0;

1 1 1];

% INPUT THE NODE COORDINATE (X,Y) MATRIX, NODECOOR (ROW # = NODE #)

nodecoor=[

0 0;

7416 3000;

15416 3000

];

% INPUT THE ELEMENT CONNECTIVITY MATRIX, LNODS (ROW # = ELEMENT #)

lnods=[

1 2;

2 3

];

% INPUT THE MATERIAL PROPERTIES ASSOCIATED WITH THIS TYPE OF STRUCTURE

% PUT INTO ARRAYS WHERE THE INDEX NUMBER IS EQUAL TO THE CORRESPONDING ELEMENT NUMBER.

% COMMENT OUT VARIABLES THAT WILL NOT BE USED

E=[200 200];

A=[6000 6000];

Iz=[200000000 200000000];

% INPUT THE LOAD DATA. NODAL LOADS, PNODS SHOULD BE IN MATRIX FORM. THE COLUMNS CORRESPOND

% TO THE GLOBAL DEGREE OF FREEDOM IN WHICH THE LOAD IS ACTING AND THE THE ROW NUMBER CORRESPONDS

% WITH THE LOAD CASE NUMBER. PELEM IS THE ELEMENT LOAD, GIVEN IN A MATRIX, WITH COLUMNS

% CORRESPONDING TO THE ELEMENT NUMBER AND ROW THE LOAD CASE. ARRAY "A" IS THE DISTANCE FROM

% THE LEFT END OF THE ELEMENT TO THE LOAD, IN ARRAY FORM. THE DISTRIBUTED LOAD, W SHOULD BE

% IN MATRIX FORM ALSO WITH COLUMNS = ELEMENT NUMBER UPON WHICH W IS ACTING AND ROWS = LOAD CASE.

% ZEROS SHOULD BE USED IN THE MATRICES WHEN THERE IS NO LOAD PRESENT. NODAL LOADS SHOULD

% BE GIVEN IN GLOBAL COORDINATES, WHEREAS THE ELEMENT LOADS AND DISTRIBUTED LOADS SHOULD BE

% GIVEN IN LOCAL COORDINATES.

Pnods=[18.75 -46.35 0];

Pelem=[0 0];

a=[0 0];

w=[0 4/1000];

% IF YOU WANT THE PROGRAM TO DRAW THE STUCTURE SET DRAWFLAG=1, IF NOT SET IT EQUAL TO 0.

% THIS IS USEFUL FOR CHECKING THE INPUT DATA.

drawflag=1;

% END OF INPUT DATA FILE

26.5.1.3 Program Implementation

In order to ”run” the program, open a new MATLAB Notebook. On the first line, type the name of themain program CASAP and evaluate that line by typing ctrl-enter. At this point, the main program readsthe input file you have just created and calls the appropriate subroutines to analyze your structure.In doing so, your input data is echoed into your MATLAB notebook and the program results are alsodisplayed. As a note, the program can also be executed directly from the MATAB workspace window,without Microsoft Word.



26.5.2 Program Listing

26.5.2.1 Main Program

%**********************************************************************************************

%Main Program: casap.m

%

% This is the main program, Computer Aided Structural Analysis Program

% CASAP. This program primarily contains logic for calling scriptfiles and does not

% perform calculations.

%

% All variables are global, but are defined in the scriptfiles in which they are used.

%

% Associated scriptfiles:

%

% (for all stuctures)

% indat.m (input data file)

% idrasmbl.m

% elmcoord.m

% draw.m

%

% (3 - for 2D-frames)

% length3.m

% stiffl3.m

% trans3.m

% assembl3.m

% loads3.m

% disp3.m

% react3.m

%

% By Dean A. Frank

% CVEN 5525

% Advanced Structural Analysis - Term Project

% Fall 1995

%



%

%**********************************************************************************************

% COMMENT CARDS ARE IN ALL CAPITALS

% SET NUMERIC FORMAT

format short e

% CLEAR MEMORY OF ALL VARIABLES

clear

% INITIALIZE OUTPUT FILE

fid = fopen(’casap.out’, ’wt’);

% SET ISTRTP EQUAL TO THE NUMBER CORRESPONDING TO THE TYPE OF STRUCTURE:

% 3 = 2DFRAME

istrtp=3;

% READ INPUT DATA SUPPLIED BY THE USER

indat

% REASSAMBLE THE ID MATRIX AND CALCULATE THE LM VECTORS

% CALL SCRIPTFILE IDRASMBL

idrasmbl

% ASSEMBLE THE ELEMENT COORDINATE MATRIX

elmcoord

% 2DFRAME CALCULATIONS



% CALCULATE THE LENGTH AND ORIENTATION ANGLE, ALPHA FOR EACH ELEMENT

% CALL SCRIPTFILE LENGTH3.M

length3

% CALCULATE THE 2DFRAME ELEMENT STIFFNESS MATRIX IN LOCAL COORDINATES

% CALL SCRIPTFILE STIFFL3.M

stiffl3

% CALCULATE THE 2DFRAME ELEMENT STIFFNESS MATRIX IN GLOBAL COORDINATES

% CALL SCRIPTFILE TRANS3.M

trans3

% ASSEMBLE THE GLOBAL STRUCTURAL STIFFNESS MATRIX

% CALL SCRIPTFILE ASSEMBL3.M

assembl3

% PRINT STRUCTURAL INFO

print_general_info

% LOOP TO PERFORM ANALYSIS FOR EACH LOAD CASE

for iload=1:nload

print_loads

% DETERMINE THE LOAD VECTOR IN GLOBAL COORDINATES

% CALL SCRIPTFILE LOADS3.M

loads3

% CALCULATE THE DISPLACEMENTS

% CALL SCRIPTFILE DISP3.M

disp3

% CALCULATE THE REACTIONS AT THE RESTRAINED DEGREES OF FREEDOM

% CALL SCRIPTFILE REACT3.M

react3

% CALCULATE THE INTERNAL FORCES FOR EACH ELEMENT

intern3

% END LOOP FOR EACH LOAD CASE

end

% DRAW THE STRUCTURE, IF USER HAS REQUESTED (DRAWFLAG=1)

% CALL SCRIPTFILE DRAW.M

draw

st=fclose(’all’);

% END OF MAIN PROGRAM (CASAP.M)

disp(’Program completed! - See "casap.out" for complete output’);

26.5.2.2 Assembly of ID Matrix

%************************************************************************************************

%SCRIPTFILE NAME: IDRASMBL.M

%

%MAIN FILE : CASAP

%

%Description : This file re-assambles the ID matrix such that the restrained

% degrees of freedom are given negative values and the unrestrained



% degrees of freedom are given incremental values begining with one

% and ending with the total number of unrestrained degrees of freedom.

%

% By Dean A. Frank

% CVEN 5525


% Fall 1995

%



%

%************************************************************************************************

% TAKE CARE OF SOME INITIAL BUSINESS: TRANSPOSE THE PNODS ARRAY

Pnods=Pnods.’;

% SET THE COUNTER TO ZERO

count=1;

negcount=-1;

% REASSEMBLE THE ID MATRIX

if istrtp==3

ndofpn=3;

nterm=6;

else

error(’Incorrect structure type specified’)

end

% SET THE ORIGINAL ID MATRIX TO TEMP MATRIX

orig_ID=ID;

% REASSEMBLE THE ID MATRIX, SUBSTITUTING RESTRAINED DEGREES OF FREEDOM WITH NEGATIVES,

% AND NUMBERING GLOBAL DEGREES OF FREEDOM

for inode=1:npoin

for icoord=1:ndofpn

if ID(inode,icoord)==0

ID(inode,icoord)=count;

count=count+1;

elseif ID(inode,icoord)==1

ID(inode,icoord)=negcount;

negcount=negcount-1;

else

error(’ID input matrix incorrect’)

end

end

end

% CREATE THE LM VECTORS FOR EACH ELEMENT

for ielem=1:nelem

LM(ielem,1:ndofpn)=ID(lnods(ielem,1),1:ndofpn);

LM(ielem,(ndofpn+1):(2*ndofpn))=ID(lnods(ielem,2),1:ndofpn);

end

% END OF IDRASMBL.M SCRIPTFILE

26.5.2.3 Element Nodal Coordinates

%**********************************************************************************************

%SCRIPTFILE NAME: ELEMCOORD.M

%

%MAIN FILE : CASAP

%

%Description : This file assembles a matrix, elemcoor which contains the coordinates

% of the first and second nodes on each element, respectively.

%



% By Dean A. Frank

% CVEN 5525


% Fall 1995

%



%

%**********************************************************************************************

% ASSEMBLE THE ELEMENT COORDINATE MATRIX, ELEMCOOR FROM NODECOOR AND LNODS

for ielem=1:nelem

elemcoor(ielem,1)=nodecoor(lnods(ielem,1),1);


%elemcoor(ielem,3)=nodecoor(lnods(ielem,1),3);



%elemcoor(ielem,6)=nodecoor(lnods(ielem,2),3);

end

% END OF ELMCOORD.M SCRIPTFILE

26.5.2.4 Element Lengths

%**********************************************************************************************

% Scriptfile name : length3.m (for 2d-frame structures)

%

% Main program : casap.m

%

% When this file is called, it computes the length of each element and the

% angle alpha between the local and global x-axes. This file can be used

% for 2-dimensional elements such as 2-D truss, 2-D frame, and grid elements.

% This information will be useful for transformation between local and global

% variables.

%

% Variable descriptions: (in the order in which they appear)

%

% nelem = number of elements in the structure

% ielem = counter for loop

% L(ielem) = length of element ielem

% elemcoor(ielem,4) = xj-coordinate of element ielem

% elemcoor(ielem,1) = xi-coordinate of element ielem

% elemcoor(ielem,5) = yj-coordinate of element ielem

% elemcoor(ielem,2) = yi-coordinate of element ielem

% alpha(ielem) = angle between local and global x-axes

%

% By Dean A. Frank


% Fall 1995

%



%

%**********************************************************************************************

% COMPUTE THE LENGTH AND ANGLE BETWEEN LOCAL AND GLOBAL X-AXES FOR EACH ELEMENT

for ielem=1:nelem

L(ielem)=

XXXXXXXXXXXXXXXXXXXXXX COMPLETE XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

alpha(ielem)=


% END OF LENGTH3.M SCRIPTFILE

26.5.2.5 Element Stiffness Matrices

%**********************************************************************************************

% Scriptfile name: stiffl3.m (for 2d-frame structures)

%



% Main program: casap.m

%

% When this file is called, it computes the element stiffenss matrix

% of a 2-D frame element in local coordinates. The element stiffness

% matrix is calculated for each element in the structure.

%

% The matrices are stored in a single matrix of dimensions 6x6*i and

% can be recalled individually later in the program.

%

% Variable descriptions: (in the order in which the appear)

%


% nelem = number of element in the structure

% k(ielem,6,6)= element stiffness matrix in local coordinates

% E(ielem) = modulus of elasticity of element ielem

% A(ielem) = cross-sectional area of element ielem

% L(ielem) = lenght of element ielem

% Iz(ielem) = moment of inertia with respect to the local z-axis of element ielem

%

% By Dean A. Frank


% Fall 1995

%

%**********************************************************************************************

for ielem=1:nelem

k(1:6,1:6,ielem)=...


end

% END OF STIFFL3.M SCRIPTFILE

26.5.2.6 Transformation Matrices

%**********************************************************************************************

% Scriptfile name : trans3.m (for 2d-frame structures)

%


%

% This file calculates the rotation matrix and the element stiffness

% matrices for each element in a 2D frame.

%


%

% ielem = counter for the loop


% rotation = rotation matrix containing all elements info

% Rot = rotational matrix for 2d-frame element

% alpha(ielem) = angle between local and global x-axes

% K = element stiffness matrix in global coordinates

% k = element stiffness matrix in local coordinates

%

% By Dean A. Frank


% Fall 1995

%

%**********************************************************************************************

% CALCULATE THE ELEMENT STIFFNESS MATRIX IN GLOBAL COORDINATES

% FOR EACH ELEMENT IN THE STRUCTURE

for ielem=1:nelem

% SET UP THE ROTATION MATRIX, ROTATAION

rotation(1:6,1:6,ielem)=...


ktemp=k(1:6,1:6,ielem);

% CALCULATE THE ELEMENT STIFFNESS MATRIX IN GLOBAL COORDINATES

Rot=rotation(1:6,1:6,ielem);

K(1:6,1:6,ielem)=




end

% END OF TRANS3.M SCRIPTFILE

26.5.2.7 Assembly of the Augmented Stiffness Matrix

%**********************************************************************************************

% Scriptfile name : assembl3.m (for 2d-frame structures)

%


%

% This file assembles the global structural stiffness matrix from the

% element stiffness matrices in global coordinates using the LM vectors.

% In addition, this file assembles the augmented stiffness matrix.

%

% Variable Descritpions (in order of appearance):

%

% ielem = Row counter for element number

% nelem = Number of elements in the structure

% iterm = Counter for term number in LM matrix

% LM(a,b) = LM matrix

% jterm = Column counter for element number

% temp1 = Temporary variable




% number_gdofs = Number of global dofs

% new_LM = LM matrix used in assembling the augmented stiffness matrix

% aug_total_dofs = Total number of structure dofs

% K_aug = Augmented structural stiffness matrix

% Ktt = Structural Stiffness Matrix (Upper left part of Augmented structural stiffness matrix)

% Ktu = Upper right part of Augmented structural stiffness matrix

% Kut = Lower left part of Augmented structural stiffness matrix

% Kuu = Lower rigth part of Augmented structural stiffness matrix

%

%

% By Dean A. Frank


% Fall 1995

%



%

%**********************************************************************************************

% RENUMBER DOF INCLUDE ALL DOF, FREE DOF FIRST, RESTRAINED NEXT

new_LM=LM;

number_gdofs=max(LM(:));

new_LM(find(LM<0))=number_gdofs-LM(find(LM<0));

aug_total_dofs=max(new_LM(:));

% ASSEMBLE THE AUGMENTED STRUCTURAL STIFFNESS MATRIX

K_aug=zeros(aug_total_dofs);

for ielem=1:nelem


Tough one!


end

% SET UP SUBMATRICES FROM THE AUGMENTED STIFFNESS MATRIX

Ktt=


Ktu=


Kut=


Kuu=


% END OF ASSEMBL3.M SCRIPTFILE



26.5.2.8 Print General Information

%**********************************************************************************************

% Scriptfile name : print_general_info.m

%


%

% Prints the general structure info to the output file

%

% By Pawel Smolarkiewicz, 3/16/99


%

%**********************************************************************************************

fprintf(fid,’\n\nNumber of Nodes: %d\n’,npoin);

fprintf(fid,’Number of Elements: %d\n’,nelem);

fprintf(fid,’Number of Load Cases: %d\n’,nload);

fprintf(fid,’Number of Restrained dofs: %d\n’,abs(min(LM(:))));

fprintf(fid,’Number of Free dofs: %d\n’,max(LM(:)));

fprintf(fid,’\nNode Info:\n’);

for inode=1:npoin

fprintf(fid,’ Node %d (%d,%d)\n’,inode,nodecoor(inode,1),nodecoor(inode,2));

freedof=’ ’;

if(ID(inode,1))>0

freedof=strcat(freedof,’ X ’);

end

if(ID(inode,2))>0

freedof=strcat(freedof,’ Y ’);

end

if(ID(inode,3))>0

freedof=strcat(freedof,’ Rot’);

end

if freedof==’ ’

freedof=’ none; node is fixed’;

end

fprintf(fid,’ Free dofs:%s\n’,freedof);

end

fprintf(fid,’\nElement Info:\n’);

for ielem=1:nelem

fprintf(fid,’ Element %d (%d->%d)’,ielem,lnods(ielem,1),lnods(ielem,2));

fprintf(fid,’ E=%d A=%d Iz=%d \n’,E(ielem),A(ielem),Iz(ielem));

end

26.5.2.9 Print Load

%**********************************************************************************************

% Scriptfile name : print_loads.m

%


%

% Prints the current load case data to the output file

%



%

%**********************************************************************************************

Load_case=iload

if iload==1

fprintf(fid,’\n_________________________________________________________________________\n\n’);

end

fprintf(fid,’Load Case: %d\n\n’,iload);

fprintf(fid,’ Nodal Loads:\n’);

for k=1:max(LM(:));

%WORK BACKWARDS WITH LM MATRIX TO FIND NODE# AND DOF

LM_spot=find(LM’==k);

elem=fix(LM_spot(1)/(nterm+1))+1;

dof=mod(LM_spot(1)-1,nterm)+1;

node=lnods(elem,fix(dof/4)+1);



switch(dof)

case {1,4}, dof=’Fx’;

case {2,5}, dof=’Fy’;

otherwise, dof=’ M’;

end

%PRINT THE DISPLACEMENTS

if Pnods(k)~=0

fprintf(fid,’ Node: %2d %s = %14d\n’,node, dof, Pnods(k));

end

end

fprintf(fid,’\n Elemental Loads:\n’);

for k=1:nelem

fprintf(fid,’ Element: %d Point load = %d at %d from left\n’,k,Pelem(k),a(k));

fprintf(fid,’ Distributed load = %d\n’,w(k));

end

fprintf(fid,’\n’);

26.5.2.10 Load Vector

%**********************************************************************************************

% Scriptfile name: loads3.m (for 2d-frame structures)

%

% Main program: casap.m

%

% When this file is called, it computes the fixed end actions for elements which

% carry distributed loads for a 2-D frame.

%


%



% b(ielem) = distance from the right end of the element to the point load

% L(ielem) = length of the element

% a(ielem) = distance from the left end of the element to the point load

% Ffl = fixed end force (reaction) at the left end due to the point load

% w(ielem) = distributed load on element ielem

% L(ielem) = length of element ielem

% Pelem(ielem) = element point load on element ielem

% Mfl = fixed end moment (reaction) at the left end due to the point load

% Ffr = fixed end force (reaction) at the right end due to the point load

% Mfr = fixed end moment (reaction) at the right end due to the point load

% feamatrix_local = matrix containing resulting fixed end actions in local coordinates

% feamatrix_global = matrix containing resulting fixed end actions in global coordinates

% fea_vector = vector of fea’s in global dofs, used to calc displacements

% fea_vector_abs = vector of fea’s in every structure dof

% dispflag = flag indicating initial displacements

% Ffld = fea (vert force) on left end of element due to initial disp

% Mfld = fea (moment) on left end of element due to initial disp

% Ffrd = fea (vert force) on right end of element due to initial disp

% Mfrd = fea (moment) on right end of element due to initial disp

% fea_vector_disp = vector of fea’s due to initial disp, used to calc displacements

% fea_vector_react = vector of fea’s due to initial disp, used to calc reactions

%

% By Dean A. Frank


% Fall 1995

%

%**********************************************************************************************

% CALCULATE THE FIXED END ACTIONS AND INSERT INTO A MATRIX IN WHICH THE ROWS CORRESPOND

% WITH THE ELEMENT NUMBER AND THE COLUMNS CORRESPOND WITH THE ELEMENT LOCAL DEGREES

% OF FREEDOM

for ielem=1:nelem

b(ielem)=L(ielem)-a(ielem);

Ffl=((w(ielem)*L(ielem))/2)+((Pelem(ielem)*(b(ielem))^2)/(L(ielem))^3)*(3*a(ielem)+b(ielem));

Mfl=((w(ielem)*(L(ielem))^2))/12+(Pelem(ielem)*a(ielem)*(b(ielem))^2)/(L(ielem))^2;

Ffr=((w(ielem)*L(ielem))/2)+((Pelem(ielem)*(a(ielem))^2)/(L(ielem))^3)*(a(ielem)+3*b(ielem));



Mfr=-((w(ielem)*(L(ielem))^2))/12+(Pelem(ielem)*a(ielem)*(b(ielem))^2)/(L(ielem))^2;

feamatrix_local(ielem,1:6)=[0 Ffl Mfl 0 Ffr Mfr];

% ROTATE THE LOCAL FEA MATRIX TO GLOBAL

feamatrix_global=...


end

% CREATE A LOAD VECTOR USING THE LM MATRIX

% INITIALIZE FEA VECTOR TO ALL ZEROS

for idofpn=1:ndofpn

fea_vector(idofpn,1)=0;

end

for ielem=1:nelem

for idof=1:6

if ielem==1

if LM(ielem,idof)>0

fea_vector(LM(ielem,idof),1)=feamatrix_global(idof,ielem);

end

elseif ielem>1

if LM(ielem,idof)>0

fea_vector(LM(ielem,idof),1)=fea_vector(LM(ielem,1))+feamatrix_global(idof,ielem);

end

end

end

end

for ielem=1:nelem

for iterm=1:nterm

if feamatrix_global(iterm,ielem)==0

else

if new_LM(ielem,iterm)>number_gdofs

fea_vector_react(iterm,1)=feamatrix_global(iterm,ielem);

end

end

end

end

% END OF LOADS3.M SCRIPTFILE

26.5.2.11 Nodal Displacements

%**********************************************************************************************

% Scriptfile name : disp3.m (for 2d-frame structures)

%


%

% When this file is called, it computes the displacements in the global

% degrees of freedom.

%


%

% Ksinv = inverse of the structural stiffness matrix

% Ktt = structural stiffness matrix

% Delta = vector of displacements for the global degrees of freedom

% Pnods = vector of nodal loads in the global degrees of freedom

% fea_vector = vector of fixed end actions in the global degrees of freedom

%

% By Dean A. Frank


% Fall 1995

%





%

%**********************************************************************************************

% CREATE A TEMPORARY VARIABLE EQUAL TO THE INVERSE OF THE STRUCTURAL STIFFNESS MATRIX

Ksinv=inv(Ktt);

% CALCULATE THE DISPLACEMENTS IN GLOBAL COORDINATES

Delta=


% PRINT DISPLACEMENTS WITH NODE INFO

fprintf(fid,’ Displacements:\n’);

for k=1:size(Delta,1)


LM_spot=find(LM’==k);




switch(dof)

case {1,4}, dof=’delta X’;

case {2,5}, dof=’delta Y’;

otherwise, dof=’rotate ’;

end

%PRINT THE DISPLACEMENTS

fprintf(fid,’ (Node: %2d %s) %14d\n’,node, dof, Delta(k));

end


% END OF DISP3.M SCRIPTFILE

26.5.2.12 Reactions

%**********************************************************************************************

% Scriptfile name : react3.m (for 2d-frame structures)

%


%

% When this file is called, it calculates the reactions at the restrained degrees of

% freedom.

%

% Variable Descriptions:

%

% Reactions = Reactions at restrained degrees of freedom

% Kut = Upper left part of aug stiffness matrix, normal structure stiff matrix

% Delta = vector of displacements

% fea_vector_react = vector of fea’s in restrained dofs

%

%

% By Dean A. Frank


% Fall 1995

%



%

%**********************************************************************************************

% CALCULATE THE REACTIONS FROM THE AUGMENTED STIFFNESS MATRIX

Reactions=


fprintf(fid,’ Reactions:\n’);

for k=1:size(Reactions,1)


LM_spot=find(LM’==-k);






switch(dof)



otherwise, dof=’M ’;

end

%PRINT THE REACTIONS

fprintf(fid,’ (Node: %2d %s) %14d\n’,node, dof, Reactions(k));

end


% END OF REACT3.M SCRIPTFILE

26.5.2.13 Internal Forces

%**********************************************************************************************

% Scriptfile name : intern3.m (for 2d-frame structures)

%


%

% When this file is called, it calculates the internal forces in all elements

% freedom.

%



%

%**********************************************************************************************

Pglobe=zeros(6,nelem);

Plocal=Pglobe;

fprintf(fid,’ Internal Forces:’);

%LOOP FOR EACH ELEMENT

for ielem=1:nelem

%FIND ALL 6 LOCAL DISPLACEMENTS

elem_delta=zeros(6,1);

for idof=1:6

gdof=LM(ielem,idof);

if gdof<0

elem_delta(idof)=


else

elem_delta(idof)=


end

end

%SOLVE FOR ELEMENT FORCES (GLOBAL)

Pglobe(:,ielem)=K(:,:,ielem)*elem_delta+feamatrix_global(:,ielem);

%ROTATE FORCES FROM GLOBAL TO LOCAL COORDINATES

%ROTATE FORCES TO LOCAL COORDINATES

Plocal(:,ielem)=


%PRINT RESULTS

fprintf(fid,’\n Element: %2d\n’,ielem);

for idof=1:6

if idof==1

fprintf(fid,’ At Node: %d\n’,lnods(ielem,1));

end

if idof==4

fprintf(fid,’ At Node: %d\n’,lnods(ielem,2));

end

switch(idof)



otherwise, dof=’M ’;

end

fprintf(fid,’ (Global : %s ) %14d’,dof, Pglobe(idof,ielem));



fprintf(fid,’ (Local : %s ) %14d\n’,dof, Plocal(idof,ielem));

end

end

fprintf(fid,’\n_________________________________________________________________________\n\n’);

26.5.2.14 Plotting

%************************************************************************************************

% SCRIPTFILE NAME : DRAW.M

%

% MAIN FILE : CASAP

%

% Description : This file will draw a 2D or 3D structure (1D structures are generally

% boring to draw).

%

% Input Variables : nodecoor - nodal coordinates

% ID - connectivity matrix

% drawflag - flag for performing drawing routine

%

% By Dean A. Frank

% CVEN 5525


% Fall 1995

%

% (with thanks to Brian Rose for help with this file)

%

%************************************************************************************************

% PERFORM OPERATIONS IN THIS FILE IF DRAWFLAG = 1

if drawflag==1

if istrtp==1

drawtype=2;

elseif istrtp==2

drawtype=2;

elseif istrtp==3

drawtype=2;

elseif istrtp==4

drawtype=2;

elseif istrtp==5

drawtype=3;

elseif istrtp==6

drawtype=3;

else

error(’Incorrect structure type in indat.m’)

end

ID=orig_ID.’;

% DRAW 2D STRUCTURE IF DRAWTYPE=2

if drawtype==2

% RETREIVE NODAL COORDINATES

x=nodecoor(:,1);

y=nodecoor(:,2);

%IF 2D-TRUSS, MODIFY ID MATRIX

if istrtp==2

for ipoin=1:npoin

if ID(1:2,ipoin)==[0;0]

ID(1:3,ipoin)=[0;0;0]

elseif ID(1:2,ipoin)==[0;1]

ID(1:3,ipoin)=[0;1;0]

elseif ID(1:2,ipoin)==[1;1]

ID(1:3,ipoin)=[1;1;0]

end



end

end

%if size(ID,1)==2

% ID=[ID;zeros(1,size(ID,2))];

%end

% IF GRID, SET ID=ZEROS

if istrtp==4

ID=ID*0;

end

% SET UP FIGURE

handle=figure;

margin=max(max(x)-min(x),max(y)-min(y))/10;

axis([min(x)-margin, max(x)+margin, min(y)-margin, max(y)+margin])

axis(’equal’)

hold on

% CALC NUMBER OF NODES, ETC.

number_nodes=length(x);

number_elements=size(lnods,1);

number_fixities=size(ID,2);

axislimits=axis;

circlesize=max(axislimits(2)-axislimits(1),axislimits(4)-axislimits(3))/40;

% DRAW SUPPORTS

for i=1:number_fixities

% DRAW HORIZ. ROLLER

if ID(:,i)==[0 1 0]’

plot(sin(0:0.1:pi*2)*circlesize/2+x(i),cos(0:0.1:pi*2)*circlesize/2-circlesize/2+y(i),’r’)

% DRAW PIN SUPPORT

elseif ID(:,i)==[1 1 0]’

plot([x(i),x(i)-circlesize,x(i)+circlesize,x(i)],[y(i),y(i)-circlesize,y(i)-circlesize,y(i)],’r’)

% DRAW HOERIZ. ROLLER SUPPORT

elseif ID(:,i)==[0 1 1]’

plot([x(i)+circlesize*2,x(i)-circlesize*2],[y(i),y(i)],’r’);

plot(sin(0:0.1:pi*2)*circlesize/2+x(i),cos(0:0.1:pi*2)*circlesize/2-circlesize/2+y(i),’r’)

plot(sin(0:0.1:pi*2)*circlesize/2+x(i)+circlesize,cos(0:0.1:pi*2)*circlesize/2-circlesize/2+y(i),’r’)

plot(sin(0:0.1:pi*2)*circlesize/2+x(i)-circlesize,cos(0:0.1:pi*2)*circlesize/2-circlesize/2+y(i),’r’)

% DRAW VERT. ROLLER SUPPORT

elseif ID(:,i)==[1 0 0]’

plot(sin(0:0.1:pi*2)*circlesize/2+x(i)-circlesize*.5,cos(0:0.1:pi*2)*circlesize/2,’r’)

% DRAW ROLLER SUPPORT WITH NO ROTATION

elseif ID(:,i)==[1 0 1]’

plot([x(i),x(i)],[y(i)+circlesize*2,y(i)-circlesize*2],’r’);

plot(sin(0:0.1:pi*2)*circlesize/2+x(i)-circlesize*.5,cos(0:0.1:pi*2)*circlesize/2,’r’)

plot(sin(0:0.1:pi*2)*circlesize/2+x(i)-circlesize*.5,cos(0:0.1:pi*2)*circlesize/2+circlesize,’r’)

plot(sin(0:0.1:pi*2)*circlesize/2+x(i)-circlesize*.5,cos(0:0.1:pi*2)*circlesize/2-circlesize,’r’)

end

% DRAW HORIZ. PLATFORM

if min(ID(:,i)==[0 1 0]’) | min(ID(:,i)==[1 1 0]’) | min(ID(:,i)==[0 1 1]’)

plot([x(i)-circlesize*2,x(i)+circlesize*2],[y(i)-circlesize,y(i)-circlesize],’r’)

plot([x(i)-circlesize*1.5,x(i)-circlesize*2],[y(i)-circlesize,y(i)-circlesize*1.5],’r’)

plot([x(i)-circlesize*1,x(i)-circlesize*2],[y(i)-circlesize,y(i)-circlesize*2],’r’)

plot([x(i)-circlesize*.5,x(i)-circlesize*1.5],[y(i)-circlesize,y(i)-circlesize*2],’r’)



plot([x(i)+circlesize*0,x(i)-circlesize*1],[y(i)-circlesize,y(i)-circlesize*2],’r’)

plot([x(i)+circlesize*.5,x(i)-circlesize*(0.5)],[y(i)-circlesize,y(i)-circlesize*2],’r’)

plot([x(i)+circlesize*1,x(i)+circlesize*0],[y(i)-circlesize,y(i)-circlesize*2],’r’)

plot([x(i)+circlesize*1.5,x(i)+circlesize*.5],[y(i)-circlesize,y(i)-circlesize*2],’r’)

plot([x(i)+circlesize*2,x(i)+circlesize*1],[y(i)-circlesize,y(i)-circlesize*2],’r’)

plot([x(i)+circlesize*2,x(i)+circlesize*1.5],[y(i)-circlesize*1.5,y(i)-circlesize*2],’r’)

% DRAW FIXED SUPPORT

elseif ID(:,i)==[1 1 1]’

plot([x(i)-circlesize*2,x(i)+circlesize*2],[y(i)-circlesize,y(i)-circlesize]+circlesize,’r’)

plot([x(i)-circlesize*1.5,x(i)-circlesize*2],[y(i)-circlesize,y(i)-circlesize*1.5]+circlesize,’r’)

plot([x(i)-circlesize*1,x(i)-circlesize*2],[y(i)-circlesize,y(i)-circlesize*2]+circlesize,’r’)

plot([x(i)-circlesize*.5,x(i)-circlesize*1.5],[y(i)-circlesize,y(i)-circlesize*2]+circlesize,’r’)

plot([x(i)+circlesize*0,x(i)-circlesize*1],[y(i)-circlesize,y(i)-circlesize*2]+circlesize,’r’)

plot([x(i)+circlesize*.5,x(i)-circlesize*(0.5)],[y(i)-circlesize,y(i)-circlesize*2]+circlesize,’r’)

plot([x(i)+circlesize*1,x(i)+circlesize*0],[y(i)-circlesize,y(i)-circlesize*2]+circlesize,’r’)

plot([x(i)+circlesize*1.5,x(i)+circlesize*.5],[y(i)-circlesize,y(i)-circlesize*2]+circlesize,’r’)

plot([x(i)+circlesize*2,x(i)+circlesize*1],[y(i)-circlesize,y(i)-circlesize*2]+circlesize,’r’)

plot([x(i)+circlesize*2,x(i)+circlesize*1.5],[y(i)-circlesize*1.5,y(i)-circlesize*2]+circlesize,’r’)

% DRAW VERT. PLATFORM

elseif min(ID(:,i)==[1 0 0]’) | min(ID(:,i)==[1 0 1]’)

xf=[x(i)-circlesize,x(i)-circlesize*2];

yf=[y(i),y(i)- circlesize];

plot(xf,yf,’r’)

plot(xf,yf+circlesize*.5,’r’)

plot(xf,yf+circlesize*1,’r’)

plot(xf,yf+circlesize*1.5,’r’)

plot(xf,yf+circlesize*2,’r’)

plot([x(i)-circlesize*1.5,x(i)-circlesize*2],[y(i)+circlesize*2,y(i)+ circlesize*1.5],’r’)

plot(xf,yf-circlesize*.5,’r’)

plot(xf,yf-circlesize*1,’r’)

plot([x(i)-circlesize,x(i)-circlesize*1.5],[y(i)-circlesize*1.5,y(i)- circlesize*2],’r’)

plot([xf(1),xf(1)],[y(i)+circlesize*2,y(i)-circlesize*2],’r’);

end

end

% DRAW ELEMENTS

for i=1:number_elements

plot([x(lnods(i,1)),x(lnods(i,2))],[y(lnods(i,1)),y(lnods(i,2))],’b’);

if i==1

end

end

% DRAW JOINTS

for i=1:number_nodes

if ~max(ID(:,i))

plot(x(i),y(i),’mo’)

end

end

% DRAW ELEMENT NUMBERS


set(handle,’DefaultTextColor’,’blue’)

text( (x(lnods(i,1))+x(lnods(i,2)))/2+circlesize,(y(lnods(i,1))+y(lnods(i,2)))/2+circlesize,int2str(i))

end

% DRAW JOINT NUMBERS


set(handle,’DefaultTextColor’,’magenta’)

text(x(i)+circlesize,y(i)+circlesize,int2str(i))

end

if exist(’filename’)

title(filename)

end



hold off

set(handle,’DefaultTextColor’,’white’)

% DRAW 3D STRUCTURE IF DRAWTYPE=3

elseif drawtype==3

% RETREIVE NODE COORIDINATES

x=nodecoor(:,1);

y=nodecoor(:,2);

z=nodecoor(:,3);

% SET UP FIGURE

handle=figure;

margin=max([max(x)-min(x),max(y)-min(y),max(z)-min(z)])/10;

axis([min(x)-margin, max(x)+margin, min(y)-margin, max(y)+margin, min(z)-margin, max(z)+margin])

axis(’equal’)

hold on

% RETREIVE NUMBER OF NODES, ETC.

number_nodes=length(x);

number_elements=size(lnods,1);

axislimits=axis;

circlesize=max([axislimits(2)-axislimits(1),axislimits(4)-axislimits(3),axislimits(6)-axislimits(5)])/40;

% DRAW ELEMENTS


plot3([x(lnods(i,1)),x(lnods(i,2))],[y(lnods(i,1)),y(lnods(i,2))],[z(lnods(i,1)),z(lnods(i,2))],’b’);

end

% DRAW JOINTS


plot3(x(i),y(i),z(i),’mo’)

end

% DRAW ELEMENT NUMBERS


set(handle,’DefaultTextColor’,’blue’)

text( (x(lnods(i,1))+x(lnods(i,2)))/2+circlesize,(y(lnods(i,1))+y(lnods(i,2)))/2+circlesize,(z(lnods(i,1))+z(lnods(i,2)

end

% DRAW JOINT NUMBERS


set(handle,’DefaultTextColor’,’magenta’)

text(x(i)+circlesize,y(i)+circlesize,z(i)+circlesize,int2str(i))

end

if exist(’filename’)

title(filename)

end

xlabel(’x’)

ylabel(’y’)

zlabel(’z’)

% DRAW GROUND

X=x;

Y=y;

Z=z;

X=axislimits(1)-margin:margin:axislimits(2)+margin;

Y=axislimits(3)-margin:margin:axislimits(4)+margin;

Z=zeros(length(X),length(Y));

mesh(X,Y,Z)



size(X)

size(Y)

size(Z)

hold off

set(handle,’DefaultTextColor’,’white’)

hAZ=uicontrol(’style’,’slider’,’position’,[.7 .95 .3 .05],’units’,’normalized’,’min’,0,’max’,360,...

’callback’,’[az,el]=view; az=get(gco,’’val’’); view(az,el);’);

hEL=uicontrol(’style’,’slider’,’position’,[.7 .89 .3 .05],’units’,’normalized’,’min’,0,’max’,180,...

’callback’,’[az,el]=view; el=get(gco,’’val’’); view(az,el);’);

hdef=uicontrol(’style’,’pushbutton’,’callback’,’view(-37.5, 30)’,’position’,[.88 .83 .12 .05],’units’,’normalized’,’Strin

%set(handle,’units’,’normalized’)

%text(.68,.95,’azimuth’)

%text(.68,.89,’elevation’)

end

end

% END OF DRAW.M SCRIPTFILE

26.5.2.15 Sample Output File

CASAP will display figure 26.15.

0 2000 4000 6000 8000 10000 12000 14000 16000−6000

−4000

−2000

0

2000

4000

6000

8000

1

2

1

2 3

Figure 26.15: Structure Plotted with CASAP

Number of Nodes: 3

Number of Elements: 2

Number of Load Cases: 1

Number of Restrained dofs: 6

Number of Free dofs: 3

Node Info:

Node 1 (0,0)

Free dofs: none; node is fixed

Node 2 (7416,3000)

Free dofs: X Y Rot

Node 3 (15416,3000)

Free dofs: none; node is fixed

Element Info:

Element 1 (1->2) E=200 A=6000 Iz=200000000

Element 2 (2->3) E=200 A=6000 Iz=200000000



_________________________________________________________________________

Load Case: 1

Nodal Loads:

Node: 2 Fx = 1.875000e+001

Node: 2 Fy = -4.635000e+001

Elemental Loads:

Element: 1 Point load = 0 at 0 from left

Distributed load = 0

Element: 2 Point load = 0 at 0 from left

Distributed load = 4.000000e-003

Displacements:

(Node: 2 delta X) 9.949820e-001

(Node: 2 delta Y) -4.981310e+000

(Node: 2 rotate ) -5.342485e-004

Reactions:

(Node: 1 Fx) 1.304973e+002

(Node: 1 Fy) 5.567659e+001

(Node: 1 M ) 1.337416e+004

(Node: 3 Fx) -1.492473e+002

(Node: 3 Fy) 2.267341e+001

(Node: 3 M ) -4.535573e+004

Internal Forces:

Element: 1

At Node: 1

(Global : Fx ) 1.304973e+002 (Local : Fx ) 1.418530e+002

(Global : Fy ) 5.567659e+001 (Local : Fy ) 2.675775e+000

(Global : M ) 1.337416e+004 (Local : M ) 1.337416e+004

At Node: 2

(Global : Fx ) -1.304973e+002 (Local : Fx ) -1.418530e+002

(Global : Fy ) -5.567659e+001 (Local : Fy ) -2.675775e+000

(Global : M ) 8.031549e+003 (Local : M ) 8.031549e+003

Element: 2

At Node: 2

(Global : Fx ) 1.492473e+002 (Local : Fx ) 1.492473e+002


(Global : M ) -8.031549e+003 (Local : M ) -8.031549e+003

At Node: 3

(Global : Fx ) -1.492473e+002 (Local : Fx ) -1.492473e+002


(Global : M ) -4.535573e+004 (Local : M ) -4.535573e+004

_________________________________________________________________________


Draft

Chapter 27

COLUMNS

27.1 Introduction

1 Columns resist a combination of axial P and flexural load M , (or M = Pe for eccentrically appliedload).

27.1.1 Types of Columns

Types of columns, Fig. 27.1

Tied column

Spiral column

Composite column

Pipe column

tie steel

main longitudinal steel reinforcement

Figure 27.1: Types of columns

2 Lateral reinforcement, Fig. 27.2

1. Restrains longitudinal steel from outward buckling

2. Restrains Poisson’s expansion of concrete

3. Acts as shear reinforcement for horizontal (wind & earthquake) load

4. Provide ductility

very important to resist earthquake load.

27.1.2 Possible Arrangement of Bars

3 Bar arrangements, Fig. 27.3

Draft27–2 COLUMNS

X

Spiraltied

P

δ

Figure 27.2: Tied vs Spiral Reinforcement

4 bars 6 bars 8 bars

10 bars

12 bars

14 bars

16 bars

Wall column

Corner column

Figure 27.3: Possible Bar arrangements


Draft27.2 Short Columns 27–3

27.2 Short Columns

27.2.1 Concentric Loading

4 No moment applied,

Elastic Behaviour

P = fcAc + fsAs= fc(Ac + nAs)

Ultimate Strength

Pd = φPnPn = .85f ′cAc + fyAs

5 note:

1. 0.85 is obtained also from test data

2. Matches with beam theory using rect. stress block

3. Provides an adequate factor of safety

27.2.2 Eccentric Columns

5 Sources of flexure, Fig. 27.4

ML MR

Pe

Figure 27.4: Sources of Bending

1. Unsymmetric moments ML �=MR

2. uncertainty of loads (must assume a minimum eccentricity)

3. unsymmetrical reinforcement

6 Types of Failure, Fig. 27.5

1. large eccentricity of load ⇒ failure by yielding of steel

2. small eccentricity of load ⇒ failure by crushing of concrete

3. balanced condition

7 Assumptions A′s = As; ρ = As

bd = A′s

bd = f ′s = fy


Draft27–4 COLUMNS

Mn

Pn

e small

e=

0

e=

Compressionfailure range

Radial lines show

Tension failure range

Load path forgivin e

e large

eb

constant e=Mn

Pn

8 M0

P0

Figure 27.5: Load moment interaction diagram

27.2.2.1 Balanced Condition

8 There is one specific eccentricity eb = MP such that failure will be triggered by simultaneous

1. steel yielding

2. concrete crushing

9 From the strain diagram (and compatibility of concrete and steel strains), Fig. 27.6

εc = .003 (27.1-a)

εy =fyEs

(27.1-b)

c =εu

εu + εyd =

.003fyEs

+ .003d (27.1-c)

(27.1-d)

Furthermore,

ε′sc− d′ =

εcc

(27.2-a)

⇒ ε′s =c− d′c

εc (27.2-b)

thus the compression steel will be yielding (i.e. ε′s = εy) for εc = .003 and d′ = 2 in if c > 6 in

10 Equilibrium:

Pn = .85f ′cab+A′sf

′y −Asfs

fs = fyAs = A′

s

a = Pn.85f ′

cb

a = β1cbcb = .003

fyEs

+.003d

Pn,b = .85β1f ′cbd

.003fyEs

+ .003(27.3)

or

Pnb = .85β1f ′cbd87, 000

fy + 87, 000(27.4)



A fs s

εs

A fs’ s

dd’

h/2

A’Ass

b

ε’s

ae

e’

0.85f’c

Pnc

A fs s

c

εsc

A fs

A fs’

y

y

Figure 27.6: Strain and Stress Diagram of a R/C Column


Draft27–6 COLUMNS

11 To obtain Mnb we take moment about centroid of tension steel As of internal forces, this must beequal and opposite to the externally applied moment, Fig. 27.6.

Mnb = Pnbeb︸︷︷︸Mext

= .85f ′cab(d−a

2) +A′

sfy(d− d′)︸︷︷︸Mint

(27.5)

12 Note: Internal moments due to Asfy and A′sfy cancel each other for symmetric columns.

27.2.2.2 Tension Failure

Case I, e is known and e > eb In this case a and Pn are unknowns, and for failure to be triggered byfy in As we must have e > eb.

Can still assume Asfy = A′sfy

ΣFy = 0⇒ Pn = .85f ′cab⇒ a =Pn.85f ′cb

(27.6-a)

ΣM = 0⇒ Pne′ = Pn(d− a2) +A′

sf′y(d− d′) (27.6-b)

Two approaches

1. Solve iteratively for those two equations

(a) Assume a (a < h2 )

(b) From strain compatibility solve for fsc, center steel stress if applicable.(c) ΣFy = 0⇒ solve for Pn(d) ΣM = 0 with respect to tensile reinforcement,⇒ solve for Pn(e) If no convergence among the two Pn, iterate by solving for a from ΣFy = 0

2. Combine them into a quadratic equation in Pn

Pn = .85f ′cbd

−ρ− (

e′

d− 1) +

√(1− e

′

d

)2+ 2ρ

[µ

(1− d

′

d

)+e′

d

] (27.7)

whereρ = As

bd = A′s

bd

µ = fy.85f ′

c

In this case, we only have two unknown, Pn and f ′s.

a = β1c (27.8-a)

fsdef= fy (27.8-b)

f ′s = εcEsc− d′c

≤ fy (27.8-c)

C = 0.85f ′cab (27.8-d)Pn = C +A′

sf′s −Asfy (27.8-e)

Mn = Ch− a

2+ a′sf

′s

(h

2− d′

)+Asfs

(d− h

2

)(27.8-f)

Note this approach is favoured when determining the interaction diagram.



27.2.2.3 Compression Failure

Case II c is known and c < cb; Pn is unknownCase I e is known and e < eb; Pn, a and fs are unknownCompression failure occurs if e′ < e′b ⇒ εu = .003, assume f ′s = fy, and fs < fyFrom geometry

c =εu

fsEs

+ εud⇒ fs = Esεu

d− cc

= Esεud− a

β1aβ1

(27.9-a)

Pn = .85f ′cab+ A′sfy −Asfs (27.9-b)

Pne′ = .85f ′cab(d−

a

2) +A′

sfy(d− d′) (27.9-c)

this would yield a cubic equation in Pn, which can be solved analytically or by iteration

1. Assume a (a h)2. Solve for ΣM = 0 with respect to tensile reinforcement & solve for Pn3. From strain compatibility solve for fs4. Check that ΣFy = 0 & solve for a5. If ai+1 �= ai go to step 2

In this case

a = β1c (27.10-a)

fs = εcEsd− cc

≤ fy (27.10-b)


≤ fy (27.10-c)

C = 0.85f ′cab (27.10-d)Pn = C +Asfs +A′

sf′s (27.10-e)

Mn = Ch− a

2+A′

sf′s

(h

2− d′

)+Asfs

(d− h

2

)(27.10-f)

27.2.3 ACI Provisions

Case II: c is known and c > cb; fs, f ′s, and Pn are unknown1.

ρmin = 1% ACI 10.9.1ρmax = 8%ρs = 0.45(AgAc − 1) f

′c

fyACI 10.5

φ = 0.7 for tied columnsφ = 0.75 for spiral columns

(27.11)

2. whereρs minimum ratio of spiral reinforcementAg gross area of sectionAc area of core

3. A minimum of 4 bars for tied circular and rect

4. A minimum of 6 bars for spirals(10.9.2)

5. φ increases linearly to 0.9 as φPn decreases from 0.10f ′cAg or φP0, whichever is smaller, to zero(ACI 9.3.2).

6. Maximum strength is 0.8φP0 for tied columns (φ = 0.7) and 0.85φP0 for spirally reinforced columns(φ = 0.75).


Draft27–8 COLUMNS

27.2.4 Interaction Diagrams

13 Each column is characterized by its own interaction diagram, Fig. 27.7

0 φMn Mn M

0.10f’cAg

eb

Pu-Mu

Pn-Mn(Mn’Pn)

(Mnb’Pnb)e min

φPn(max)

Pn(max)

P0

P

A

Tied: Pn(max)=0.80P0

Spirally reinf: Pn(max)=0.85P0

1e

Compressioncontrol region

Tensioncontrol region

Balanced failure

Figure 27.7: Column Interaction Diagram

27.2.5 Design Charts

14 To assist in the design of R.C. columns, design charts have been generated by ACI in term of nondimensionalized parameters χ = Pn

bhf ′cvs Mn

bh2f ′c

= χeh for various ρt where ρt = As+A

′s

bh and µ = fy.85f ′

c

Example 27-1: R/C Column, c known

A 12 by 20 in. column is reinforced with four No. 4 bars of area 1.0 in2 each, at each corner.f ′c = 3.5 ksi, fy = 50 ksi, d′ = 2.5 in. Determne: 1) Pb and Mb; 2) The load and moment for c = 5 in; 3)load and moment for c = 18 in.Solution:

Balanced Conditions is derived by revisiting the fundamental equations, rather than mere substitu-tion into previously derived equation.

d = h− d′ = 20− 2.5 = 17.5 in (27.12-a)

cb =.003

fyEs

+ .003d =

.00350

29,000 + .00317.5 = 11.1 in (27.12-b)

a = β1cb = (0.85)(11.1) = 9.44 in (27.12-c)

fsdef= fy = 50 ksi (27.12-d)

f ′s = Esc− d′c

εc = (29, 000)(11.1− 2.5

11.1(0.003) = 67.4 ksi > fy ⇒ f ′s = 50 ksi(27.12-e)

C = 0.85f ′cab = (0.85)(3.5)(9.44)(12) = 337 k (27.12-f)

Pnb = C +Asfs −A′sfs = 337 + (2.0)(50) + (2.0)(−50) = 337 k (27.12-g)

Mnb = Pnbe′ = .85f ′cab(d−

a

2) +A′

sfy(d− d′) (27.12-h)

= 337(17.5− 9.44

2

)+ (2.0)(50)(17.5− 2.5) = 3, 280 k.in = 273 k.ft (27.12-i)



eb =3, 280337

= 9.72 in (27.12-j)

Tension failure, c = 5 in

fsdef= fy = 50 ksi (27.13-a)


≤ fy (27.13-b)

= (0.003)(29, 000)5.0− 2.5

5.0= 43.5 ksi (27.13-c)

a = β1c = 0.85(5.0) = 4.25 in (27.13-d)C = 0.85f ′cab (27.13-e)

= (0.85)(3.5)(4.25)(12) = 152 k (27.13-f)Pn = C +A′

sf′s −Asfy (27.13-g)

= 152 + (2.0)(43.5)− (2.0)(50) = 139 k (27.13-h)

Mn = Ch− a

2+ a′sf

′s

(h

2− d′

)+Asfs

(d− h

2

)about section centroid (27.13-i)

= (152)(

20− 4.252

)+ (2.0)(43.5)

(202− 2.5

)+ (2.0)(50)

(17.5− 20

2

)(27.13-j)

= 2, 598 k.in = 217 k.ft (27.13-k)

e =2, 598139

= 18.69 in (27.13-l)

Compression failure, c = 18 in

a = β1c = 0.85(18) = 15.3 in (27.14-a)

fs = εcEsd− cc

≤ fy (27.14-b)

= (0.003)(29, 000)17.5− 18.0

18.0= −2 ksi As is under compression (27.14-c)


≤ fy (27.14-d)

= (0.003)(29, 000)18.0− 2.5

18.0= 75 ksi > fy ⇒ f ′s = 50 ksi (27.14-e)

C = 0.85f ′cab = (0.85)(3.5)(15.3)(12) = 546 k (27.14-f)Pn = C +A′

sf′s +Asfs (27.14-g)

= 546 + (2.0)(50) + (−2.0)(2) = 650 k (27.14-h)

Mn = Ch− a

2+A′

sf′s

(h

2− d′

)+Asfs

(d− h

2

)about section centroid (27.14-i)

= (546)(

20− 15.32

)+ (2.0)(50)

(202− 2.5

)+ (2.0)(−2.0)

(17.5− 20

2

)(27.14-j)

= 2, 000 k.in = 167 k.ft (27.14-k)

e =2, 000650

= 3.07 in (27.14-l)

Example 27-2: R/C Column, e known


Draft27–10 COLUMNS

For the following column, determine eb, Pb, Mb f′c = 3, 000 psi and fy = 40, 000 psi. The area of each

bar is 1.56 in2.

3" 3"

24"

20"

12"

y

Cc

.003

cε

Balanced Condition:

εy =fyEs

=40

29, 000= .001379 (27.15-a)

cb =εu

εu + εyd =

(.003)(21).003 + .001379

= 14.4 in (27.15-b)

a = β1cb = (.85)(14.4) = 12.2 in (27.15-c)Cc = .85f ′cab = (.85)(3)(12.2)(20) = 624 k (27.15-d)

εsc =.00314.4

(14.4− 12) = .0005 (27.15-e)

fsc = (29, 000)(.0005) = 15 ksi center bars (27.15-f)Cs = (.0005)(29, 000)(2)(1.56) = 45.2 k (27.15-g)

Pnb = 624 + 45.2 = 671 k (27.15-h)

assuming that both edge bars are yielding and cancel each other.

Alternatively

Pnb = 624 + (2× 1.56)[15− (.85)(3)]− (4× 1.56)(.85)(3) = 647 k (27.16)

Note the difference 671−647647 × 100 = 3.7%

Taking moment about centroid of section

Mnb = Pnbe (27.17-a)

= .85f ′cab(12− a

2

)+Ascfy(9) + A′

s(fy − .85f ′c)(12− d′) (27.17-b)

= (.85)(3)(12.2)(20)(

12− 12.22

)+ 4(1.56)(9)(40)

+4(1.56)(40− .85× 3)(12− 3) (27.17-c)= 3, 671 + 2, 246 + 2, 103 (27.17-d)

= 8, 020 k.in; 668 k.ft (27.17-e)

eb =8, 020647

= 12.4 in (27.17-f)

e= .1 h e = (.1)(24) = 2.4 in < eb ⇒ failure by compression. Pn, a and fs are unknown



23.5

.003h-c-d’

12

εs

1. Assume a = 20 in

c =a

β1=

20.85

= 23.5 in (27.18)

2. For center steel (from geometry)

εsc

c− h2

=.003c

(27.19-a)

⇒ εsc =c− h

2

c.003 (27.19-b)

fsc = Esεsc (27.19-c)

= Esc− h

2

c.003 (27.19-d)

= 29, 00023.5− 12

23.5.003 = 42.5 ksi > fy ⇒ fsc = fy (27.19-e)

3. Take moment about centroid of tensile steel bar

Pne′ = 0.85f ′cab(d−

a

2) +A′

sfy(h− 2d′) +Ascfy(h

2− d′) (27.20-a)

Pn(9 + 2.4) = (.85)(3)(20)(20)(21− 202

) + 4(1.56)(40)(24− 6) + 2(1.56)(40)(9)(27.20-b)

= 11, 220 + 4, 493 + 259.7 (27.20-c)⇒ Pn = 1, 476 k (27.20-d)

4. Get εs in tension bar

εsh− d′ − 23.5

=.003c

(27.21-a)

⇒ εs =.00323.5

(24− 3− 23.5) (27.21-b)

= −.000319 (27.21-c)fs = Eεs = (29, 000)(0.000319) = 9.25 ksi (27.21-d)

5. Take ΣF = 0 to check assumption of a

Pn = 0.85f ′cab+A′sfy +Ascfsc +Asfs (27.22-a)

1, 476 k = (.85)(3)(a)(20) + (6)(1.56)(40) + (4)(1.56)(9.25) (27.22-b)1, 476 = 51a+ 374.4 + 57.7 (27.22-c)⇒ a = 20.4 in

√(27.22-d)

Pn = 1, 476 k (27.22-e)

Mn = (1, 476)(2.4) = 3, 542 k.in = 295 k.ft (27.22-f)

e=h 1. In this case e = 24 in > eb ⇒ failure by tension. Pn and a are unknown.



2. Assume a = 7.9 in⇒ c = aβ1

= 7.9.85 = 9.3 in

3. Steel stress at centroid

c

.003=

12− cεsc

(27.23-a)

⇒ εsc =12− 9.3

9.3.003 = .00087 (27.23-b)

⇒ fsc = (29, 000)(0.00087) = 25.3 ksi (27.23-c)

4. Iterate

ΣF = 0⇒ Pn = (.85)f ′cab+Ascfsc (27.24-a)= (.85)(3)(7.9)(20)− 2(1.56)(25.3) (27.24-b)= 403− 79 = 324 k (27.24-c)

ΣM = 0⇒ Pn(e+ h/2− d′) = .85f ′cab(d−a

2) +A′

sfy(d− d′)

−Ascfsc(d− d′

2) (27.24-d)

Pn(24 + 9) = (.85)(3)(7.9)(20)(21− 7.92

) + 4(1.56)(40)(21− 3)

+2(1.56)(25.3)(9) (27.24-e)Pn(33) = 6, 870 + 4, 493− 710 = 10, 653 k.in (27.24-f)

⇒ Pn = 323 k√

(27.24-g)

5. Determine Mn

Mn = Pne = (323)(24) = 7, 752 k.in = 646 k.ft (27.25)

% e is known solve for P_n and M_n

%

% input data

b=20;h=24;A_s=4*1.56; Ap_s=A_s;A_sc=A_s/2;fp_c=3;f_y=40;E_s=29000;dp=3;

d=h-dp;

% solve for balanced conditions first

epsi_y=f_y/E_s;

c_b=0.003*d/(0.003+epsi_y);

a=0.85*c_b;

C_c=0.85*fp_c*a*b;

epsi_sc=0.003*(c_b-h/2)/c_b;

f_sc=E_s*epsi_sc;

C_s=epsi_sc*E_s*A_s;

P_nb=C_c+A_sc*f_sc

M_nb=0.85*fp_c*a*b*(h/2-a/2)+Ap_s*f_y*(h/2-dp)+Ap_s*(f_y-0.85*fp_c)*(h/2-dp)

e_b=M_nb/P_nb

%

% e=0.1h failure by compression; assume a=20

%

e=0.1*h;

error=1.;

a=17;

iter=0;

while abs(error) > 0.1

iter=iter+1;

c=a/0.85;

epsi_sc=0.003*(c-h/2)/c ;

f_sc=E_s*epsi_sc;

if(f_sc>f_y)

f_sc=f_y;

end

M_n=0.85*fp_c*a*b*(d-a/2)+Ap_s*f_y*(h-2*dp)+A_sc*f_y*(h/2-dp);

P_n=M_n/(h/2-dp+e);

epsi_s=0.003*(h-dp-c)/c;

f_s=abs(E_s*epsi_s);

a_new=(P_n-Ap_s*f_y-A_sc*f_sc-A_s*f_s)/(0.85*fp_c*b);

error=(a_new-a)/(a_new);



a=a_new;

end

iter

P_n

M_n=P_n*e

%

% e=h; failure by tension

%

e=h;

error=1.;

a=10;

iter=0;

while abs(error) > 0.1

iter=iter+1;

c=a/0.85

epsi_sc=0.003*(c-h/2)/c ;

f_sc=E_s*epsi_sc;

if(f_sc>f_y)

f_sc=f_y;

end

P_n_1=0.85*fp_c*a*b+A_sc*f_sc

P_n_2=(0.85*fp_c*a*b*(d-a/2)+Ap_s*f_y*(d-dp)+A_sc*f_sc*(d-dp)/2)/(e+h/2-dp);

a_new=(P_n_2-A_sc*f_sc)/(0.85*fp_c*b);

error=(a_new-a)/a_new;

a=a_new;

end

iter

P_n_1

M_n=P_n_1*e

%================================

Balanced Conditions

d=21

epsi_y=0.0014

c_b=14.3858

a=12.2280

C_c=623.6256

epsi_sc=4.9754e-04

f_sc= 14.4286

C_s= 90.0343

P_nb= 668.6427

M_nb= 8.0203e+03

e_b= 11.9948

Failure by Compression

e=2.4000

error=1

a=17

iter=0

iter=1

c=20

epsi_sc=0.0012 05

f_sc=34.8000

M_n=1.6454e+04

P_n=1.4433e+03

epsi_s=1.5000e-04

f_s=4.3500

a_new=20.7445

error=0.1805

a= 20.7445

iter= 2

c=24.4053

epsi_sc=0.0015

f_sc=44.2224

f_sc=40

M_n=1.6860e+04

P_n=1.4789e+03

epsi_s=-4.1859e-04

f_s=12.1392

Failure by Tension

e=24

error=1



a=10

iter=0

iter=1

c=11.7647

epsi_sc=-6.0000e-

f_sc=-1.7400

P_n_1=504.5712

P_n_2=381.9376

a_new=7.5954

error=-0.3166

a=7.5954

iter=2

c=8.9358

epsi_sc=-0.0010

f_sc=-29.8336

P_n_1=294.2857

P_n_2=312.6867

a_new=7.9562

error=0.0453

a=7.9562

iter=2

P_n_1=294.2857

Example 27-3: R/C Column, Using Design Charts

Design the reinforcement for a column with h = 20 in, b = 12 in, d′ = 2.5 in, f ′c = 4, 000 psi,fy = 60, 000 psi, to support PDL = 56 k, PLL = 72 k, MDL = 88 k.ft, MLL = 75 k.ft,Solution:

1. Ultimate loads

Pu = (1.4)(56) + (1.7)(72) = 201 k ⇒ Pn = 2010.7 = 287 k

Mu = (1.4)(88) + (1.7)(75) = 251 k.ft ⇒ Mn = 2510.7 = 358 k.ft

(27.26)

2. Chart parameters

e

h=

(358)(12)(287)(20)

= 0.75 (27.27-a)

γ =20− (2)(2.5)

20= 0.75⇒ interpolate between A3 and A4 (27.27-b)

κ =Pnbhf ′c

=287

(12)(20)(4)= 0.3 (27.27-c)

κe

h= (0.3)(0.75) = 0.225 (27.27-d)

3. Interpolating between A.3 and A.4 ⇒ ρtµ = 0.4

4. Reinforcement

ρt = Atbh

µ = fy.85f ′

c

}At =

(0.4)(b)(h)(.85)(f ′c)fy

= (0.4)(12)(20)(.85)(4)1

(60)= 5.45 in2 (27.28-a)

(27.28-b)

⇒ use 4 # 9 & 2 # 8, ⇒ At = 5.57 in2


Draft

Chapter 28

DESIGN II

28.1 Frames

28.1.1 Beam Column Connections

1 The connection between the beam and the column can be, Fig. 28.1:

θ b

θc

θ b θcθ b θc=θ b θc=θ b θc=

θ

θ

θ

θ

b b

cc

M=K( - )s s

Flexible Rigid Semi-Flexible

Figure 28.1: Flexible, Rigid, and Semi-Flexible Joints

Flexible that is a hinge which can transfer forces only. In this case we really have cantiliver actiononly. In a flexible connection the column and beam end moments are both equal to zero, Mcol =Mbeam = 0. The end rotation are not equal, θcol �= θbeam.

Rigid: The connection is such that θbeam = θcol and moment can be transmitted through the connection.In a rigid connection, the end moments and rotations are equal (unless there is an externally appliedmoment at the node), Mcol =Mbeam �= 0, θcol = θbeam.

Semi-Rigid: The end moments are equal and not equal to zero, but the rotation are different. θbeam �=θcol, Mcol =Mbeam �= 0. Furthermore, the difference in rotation is resisted by the springMspring =Kspring(θcol − θbeam).

28.1.2 Behavior of Simple Frames

2 For vertical load across the beam rigid connection will reduce the maximum moment in the beam (atthe expense of a negative moment at the ends which will in turn be transferred to the columns).

3 The advantages of a rigid connection are greater when the frame is subjected to a lateral load. Underthose conditions, the connection will stiffen the structure and reduce the amount of lateral deflection,Fig. 28.2.

4 Fig. 28.3 illustrates the deformation, shear, moment and axial forces in frames with different boundaryconditions under both vertical and horizontal loads.

Draft28–2 DESIGN II

PI PI

PI

PI PI

PI

V V V V

H∆

H∆

θ

PI PI

θ

Figure 28.2: Deformation of Flexible and Rigid Frames Subjected to Vertical and Horizontal Loads, (Linand Stotesbury 1981)

28.1.3 Eccentricity of Applied Loads

5 A concentric axial force P and moment M , applied on a support sytem (foundation, columns, pre-stressing) can be replaced by a static equivalent one in which the moment M is eliminated and the forceP applied with an eccentricity

e =M

P(28.1)

6 The induced stresses can be decomposed into uniform (−P/L) (assuming a unit width) and linearlyvarying one (σ =M/S) and the end stresses are

σmin = −PL− σ (28.2-a)

σmax = −PL

+ σ (28.2-b)

We note that the linearly varying stress distribution must satisfy two equilibrium requirements: ΣF = 0,thus the neutral axis (where the stress is equal to zero) passes through the centroid of the section, andΣM = 0, i.e. Mint =Mext.

7 If we seek the eccentricity ecr for which σmax equals zero, then σ = PL

8 The net tensile force due to the eccentric load is

T =12σL

2(28.3)

If we want this net tensile force to be equal and opposite to the compressive force, then

T = σL4

σ = PA

}T = P

L

applied at23L2 from the centroid

(28.4)

Thus the net internal moment is

Mint = 2T23L

2= 2

P

423L

2=PL

6(28.5)


Draft28.1 Frames 28–3

w

P

w/2

-w/2

w/2

-w/2

w/2

-w/2

w/2

M’

M’

M’/

L

-M’/

L

M’/

L

-M’/

L

a

bh

L

c

d

e

f

g

h

i

j

k

l

p

M

w/2 w/2

M’

-M’/L

p

-M’/L

p

w/2 -w/2

M/h-M/h

p/2p/2

-M/L

w/2 -w/2

0.36

M/h

-0.3

6M/h

-M/L

p/2p/2

-w/2

0.68

M/h

-0.5M’/L

p/2 p/2

M

M

w/2 w/2

w/2 w/2

M M

M/h

w/2 w/2

M’/2 M’/2p/2

M/L

-M/L

0.4M 0.4M

0.64M0.4M/h

w/2 w/2

M’/2p/2

-M’/

L

M’/

L

0.55M

0.45M

M’/4M’/4

M’/4

0.68M/h

p/2

w/2w/2

M’/

2L

-M’/

2L

POST AND BEAM STRUCTURE

SIMPLE BENT FRAME

THREE-HINGE PORTAL

THREE-HINGE PORTAL

TWO-HINGE FRAME

RIGID FRAME

-0.6

8M/h

M’/4

0.45M

M’/2

Deformation Shear Moment AxialFrame TypeW=wL, M=wL/8, M’=Ph

2

Figure 28.3: Deformation, Shear, Moment, and Axial Diagrams for Various Types of Portal FramesSubjected to Vertical and Horizontal Loads



9 To satisfy the equilibrium equation, this internal moment must be equal and opposite to the externalmoment Mext = Pecr hence

PL

6︸︷︷︸Mint

= Pecr︸︷︷︸Mext

(28.6)

or

ecr =L

6(28.7)

in other words to avoid tensile stresses on either side, the resultant force P must be placed within themidle third kernel, Fig. 28.4

L/2L/2 L/3 L/3 L/3

e

L/3 L/3 L/3

e

L/3 L/3L/6

P

L/3 L/3L/6

P

L/3L/3L/2L/2

L/3L/3P PL/6

M/S

+ +

=

+

==

P/A

P

Figure 28.4: Axial and Flexural Stresses

10 This equation is fundamental in preventing tensile forces in

1. Prestressed concrete beams: If the prestressing cable is within the kernel (i.e middle third), thenthere will not be any tensile stresses caused by prestressing alone.

2. Foundations: If the eccentricity is within the middle kernel, then we have compressive stresses onlyunder the foundation and no undesirable uplift.

3. Buildings: If the eccentricity of the vertical load is within the middle third, all columns will beloaded under compression only.

28.1.4 Design of a Statically Indeterminate Arch

Adapted from (Kinney 1957)Design a two-hinged, solid welded-steel arch rib for a hangar. The moment of inertia of the rib is to

vary as necessary. The span, center to center of hinges, is to be 200 ft. Ribs are to be placed 35 ft centerto center, with a rise of 35 ft. Roof deck, purlins and rib will be assumed to weight 25 lb/ft2 on roofsurface, and snow will be assumed at 40 lb/ft2 of this surface. Twenty purlins will be equally spacedaround the rib.

1. The center line of the rib will be taken as the segment of a circle. By computation the radius ofthis circle is found to be 160.357 ft, and the length of the arc AB to be 107.984 ft.

2. For the analysis the arc AB will be considered to be divided into ten segments, each with a lengthof 10.798 ft. Thus a concentrated load is applied to the rib by the purlins framing at the center ofeach segment. (The numbered segments are indicated in Fig. ??.

3. Since the total dead and snow load is 65 lb/ft2 of roof surface, the value of each concentrated forcewill be

P = 10.798× 35× 65 = 24.565 k (28.8)



Figure 28.5: Design of a Statically Indeterminate Arch

4. The computations necessary to evaluate the coodinates of the centers of the various segments,referred to the hinge at A, are shown in Table 28.1. Also shown are the values of ∆x, the horizontalprojection of the distance between the centers of the several segments.

5. If experience is lacking and the designing engineer is therefore at a loss as to the initial assumptionsregarding the sectional variation along the rib, it is recommended that the first analysis be basedon the assumption of a constant moment of inertia. Even the experienced engineer will often findthat this is the best procedure.

6. The value of H having thus been determined, moments and axial thrusts are computed throughoutthe rib. Cross-sectional areas at the various sections are then designed for these moments and axialthrusts. A new value for H , as well as revised values for moments and axial thrust throughout therib, are determined for the new rib with the varying cross section. Stresses are checked in the rib,and the whole process is repeated if necessary. This procedure will be followed in the present case.

7. By the general method, assuming that end A is on rollers,

∆Ah +HAδAhAh = 0 (28.9)

8. Deflections will be computed by virtual work, and thus this expression becomes

∑δM

M

EI∆S +HA

∑δM

M

EI∆S = 0 (28.10)

9. Since E and ∆S are constant and, for the first analysis, the moment of inertia is constant, thefinal expression for HA is

HA = −∑MδM∑δMM

(28.11)

All the computations necessary for the first evaluation of HA are shown in Table 28.2 yielding

HA = −2∑MδM

2δMM= −2× 2, 158, 100

2× 6461.9= −333.97k (28.12)

10. The moments at the centers of the various segments of the rib, using the approximate value justdetermined for HA, are computed in Table 28.3, where a moment is considered to be positive if ittends to cause compression on the upper surface of the rib.



x′ = y′ = x = y =Segment α sin α cos α R sin α R cos α 100 − x′ y′ − 125.36 ∆x

(ft) (ft) (ft) (ft) (ft)

A 38◦34′49′′ 0.62361 0.78173 100.000 125.36 0 01 36◦39′05′′ 0.59695 0.80229 95.725 128.65 4.275 3.29 4.2752 32◦47′36′′ 0.54161 0.84063 86.851 134.80 13.149 9.44 8.8743 28◦56′07′′ 0.48382 0.87516 77.584 140.34 22.416 14.98 9.2674 25◦04′38′′ 0.42384 0.90574 67.966 145.24 32.034 19.88 9.6185 21◦13′09′′ 0.36194 0.93220 58.040 149.48 41.960 24.13 9.9266 17◦21′40′′ 0.29840 0.95444 47.851 153.05 52.149 27.69 10.1897 13◦30′11′′ 0.23350 0.97236 37.443 155.92 62.557 30.57 10.4088 09◦38′42′′ 0.16754 0.98586 26.866 158.09 73.134 32.73 10.5779 05◦47′13′′ 0.10083 0.99490 16.169 159.54 83.831 34.18 10.69710 01◦55′44′′ 0.03367 0.99943 5.399 160.27 94.601 34.91 10.770

Crown 160.36 100.000 35.00 5.399

Table 28.1: Geometry of the Arch

(2) (3) (4) (5) (6) (7) (8) (9)Shear M

to M simpleSegment x y = δM right ∆x increment beam MδM δMM

(ft) (ft · k) (k) (ft) (ft · k) (ft· k)A 0.0 0.0 245.6501 4.275 3.29 221.085 4.275 1,050 1,050 3,500 10.92 13.149 9.44 196.520 8.874 1,962 3,010 28,400 89.23 22.416 14.98 171.955 9.267 1,821 4,830 72,400 224.44 32.034 19.88 147.390 9.618 1,654 6,490 129,100 395.45 41.960 24.13 122.825 9.926 1,463 7,950 191,800 582.26 52.149 27.69 98.260 10.189 1,251 9,200 254,800 767.07 62.557 30.57 73.695 10.408 1,023 10,220 312,400 934.48 73.134 32.73 49.130 10.577 779 11,000 360,000 1,071.49 83.831 34.18 24.565 10.697 526 11,530 394,100 1,168.410 94.601 34.91 0.000 10.770 265 11,790 411,600 1,218.6

Crown 100.00 35.00 5.399 0 11,790∑2,158,100 6,461.9

Table 28.2: Calculation of Horizontal Force



Msimple Total M

Segment y beam HAy at segment(ft) (ft · k) (ft · k) (ft · k)

A 01 3.29 1,050 –1,100 –502 9.44 3,010 –3,150 –1403 14.98 4,830 –5,000 –1704 19.88 6,490 –6,640 –1505 24.13 7,950 –8,060 –1106 27.69 9,200 –9,250 –507 30.57 10,220 –10,210 +108 32.73 11,000 –10,930 +709 34.18 11,530 –11,420 +11010 34.19 11,790 –11,660 +130

Crown 35.00 11,790 –11,690 +100

Table 28.3: Moment at the Centers of the Ribs

11. The thrust N , normal to the cross section at the center of each segment, must be computed beforethe section requirements can be determined. As a matter of interest the various values of S, thetotal shear on the cross section, will also be computed for the center of each segment.

12. N and S are obtained by combining components of H and the vertical shear V at the center ofeach segment, a combination which is shown in Fig. 28.6 From this figure it is apparent that

α

αH

V sin

H cos H sin

α

αα

V c

osα

V α

Figure 28.6: Normal and Shear Forces

S = V cosα−H sinα (28.13-a)N = V sinα+G cosα (28.13-b)

13. Values of S andN are computed in Table . The values of S are included in support of the statementpreviously made to the effect that shearing forces are small in the usual arch rib.

14. Inspection of the values for the moment M in Table 28.3 and for the thrust N in Table 28.3 willindicate that segment 3 is critical in the region of negative moment and segment 10 in the region ofpositive moment. The necessary sections for the rib, at the centers of these two segments, must bedetermined in accordance with the specifications of the American Institute of Steel Constructionbefore Table 28.5 can be completed.

15. Flanges are 12 in. × 1 in., and the allowable stress for bending is FB=20 k/in2. The web thicknessis 3

4 in., and the allowable axial stress1 is FA=17,000-0.485 L/r.1In this equation the value of L is 10.798, since ribs are laterally braced at purlin framing points.



Segment V (k) V cos α − H sin α = S (k) V sin α +H cos α = N (k)A 245.6 192 -208 = -16 153 + 261 = 4141 221.1 177 -199 = -22 132 +268 = 4002 196.5 165 -181 = -16 106 +281 = 3873 172.0 150 -162 = -12 83 +292 = 3754 147.4 133 -142 = -9 62 +303 = 3655 122.8 114 -121 = -7 44 +311 = 3556 98.3 94 -100 = -6 29 +319 = 3487 73.7 72 -78 = -6 17 +325 = 3428 49.1 48 -56 = -8 8 +329 = 3379 24.6 244 -34 = -10 2 +332 = 33410 0.0 0 -11 = -11 0 +334 = 334

Crown +334 = 334

Table 28.4: Values of Normal and Shear Forces

16. Thickness of flange and web plates and width of web plates are matters of judgment. The totaldepth determined for the center of segment 3 is used from the center of this segment to the end ofthe rib. The depth determined for the center of segment 10 is arbitrarily used for the rib depth atthe crown. The total depth of the rib from the center of segment 3 to the crown is made to varylinerarly. The adequacies of the sections thus determined for the centers of the several segmentsare checked in Table 28.5.

17. It is necessary to recompute the value of HA because the rib now has a varying moment of inertia.Equation 28.11 must be altered to include the I of each segment and is now written as

HA = −∑MδM/I∑δMM/I

(28.14)

18. The revised value for HA is easily determined as shown in Table 28.6. Note that the values incolumn (2) are found by dividing the values ofMδM for the corresponding segments in column (8)of Table 28.2 by the total I for each segment as shown in Table 28.5. The values in column (3) ofTable 28.6 are found in a similar manner from the values in column (9) of Table 28.2. The simplebeam moments in column (5) of Table 28.6 are taken directly from column (7) of Table 28.2.

19. Thus the revised HA is

HA = −2∑MδM/I

2∑δMM/I

= −frac2× 1, 010.852× 3.0192 = −334.81k (28.15)

20. The revised values for the axial thrust N at the centers of the various segments are computed inTable 28.7.

21. The sections previously designed at the centers of the segments are checked for adequacy in Table28.8. From this table it appears that all sections of the rib are satisfactory. This cannot bedefinitely concluded, however, until the secondary stresses caused by the deflection of the rib areinvestigated.

28.1.5 Temperature Changes in an Arch

Adapted from (Kinney 1957)Determine the effects of rib shortening and temperature changes in the arch rib of Fig. 28.5. Consider

a temperature drop of 100◦F.

1. In order to determine the effects of rib shortening it is first necessary to find the horizontal dis-placement of end A which will result if the elastic axial strains, which occur in the actual loadedarch, are introduced into the rib while end A is on rollers. It is then necessary to find the horizontal



(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

(10)

(11)

(12)

f a=

f b=

Tot

alA

xial

N AMc

IA

llowab

leD

ecim

alde

pth

Are

aI

Mth

rustN

Fa

fa

Fa

+fb

Fb

=of

Segm

ent

(in.

)(in2

)(in4

)(ft·k

)(k

)(k

/in2

)(k

/in2

)(k

/in2

)ca

paci

tyA

23.0

039

.75

3483

414

10.4

15.8

80.

660.

661

23.0

039

.75

3483

5040

010

.12.

015

.88

0.63

0.10

0.73

223

.00

39.7

534

8314

038

79.

75.

915

.88

0.61

0.28

0.89

323

.00

39.7

534

8317

037

59.

46.

815

.88

0.60

0.34

0.94

422

.07

39.0

531

7215

036

59.

36.

315

.90

0.59

0.32

0.91

521

.13

38.3

528

6511

035

59.

24.

915

.92

0.58

0.24

0.82

620

.20

37.6

525

8550

348

9.2

2.3

15.9

40.

580.

110.

707

19.2

736

.95

2329

1034

29.

30.

515

.96

0.58

0.02

0.61

818

.32

36.2

420

7170

337

9.3

3.7

15.9

70.

580.

190.

779

17.4

035

.55

1838

110

334

9.4

6.3

15.9

90.

590.

310.

9010

16.4

734

.85

1631

130

334

9.6

7.9

16.0

10.

600.

401.

00C

row

n16

.00

34.5

015

2710

033

49.

76.

316

.03

0.60

0.32

0.92

Tab

le28

.5:D

esig

nof

the

Arc

h



(1) (2) (3) (4) (5) (6)M

Revised simple RevisedSegment Mn

Im2

I HAy beam moment(ft · k) (ft · k) (ft · k)

1 1.00 0.0031 1,100 1,050 -502 8.16 0.0256 3,160 3,010 -1503 20.79 0.0644 5,020 4,830 -1904 40.67 0.1247 6,660 6,490 -1705 66.95 0.2032 8,080 7,950 -1306 98.57 0.2967 9,270 9,200 -707 134.13 0.4012 10,240 10,220 -208 173.82 0.5174 10,960 11,000 +409 214.41 0.6357 11,450 11,530 +8010 252.35 0.7472 11,690 11,790 +100

Crown 11,720 11,790 +70∑1010.85 3.0192

Table 28.6:

RevisedSegment V sin α + H cos α = N

(k)A 153 262 4151 132 269 4012 106 282 3883 83 293 3764 62 303 3655 44 312 3566 29 320 3497 17 326 3438 8 330 3389 2 333 33510 0 335 335

Crown 335 335

Table 28.7:

NA

Mc

I DecimalN/AFA

+ Mc/IFB

= ofSegment (k/in2) (k/in2) capacityA 10.4 0.66 0.661 10.1 2.0 0.64 0.11 0.752 9.8 6.0 0.62 0.30 0.923 9.5 7.6 0.60 0.38 0.984 9.4 7.1 0.59 0.36 0.955 9.3 5.7 0.58 0.29 0.876 9,3 3.3 0.58 0.17 0.757 9.3 1.0 0.58 0.05 0.638 9.3 2.1 0.58 0.11 0.699 9.4 4.5 0.59 0.23 0.8210 9.6 6.1 0.60 0.31 0.91

Crown 9.7 4.4 0.61 0.23 0.84

Table 28.8:



N δN = cosα A NδNA

δN2

ASegment (k) (k) (in2)

1 401 0.8023 39.75 8.07 0.01622 388 0.8406 39.75 8.21 0.01783 376 0.8752 39.75 8.30 0.01934 365 0.9057 39.05 8.49 0.02105 356 0.9322 38.35 8.68 0.02276 349 0.9544 37.65 8.86 0.02427 343 0.9724 36.95 9.04 0.02568 338 0.9859 36.24 9.22 0.02689 335 0.9949 35.55 9.41 0.028610 335 0.9994 34.85 9.60 0.0278∑

87.88 0.2300

Table 28.9:

displacement of end A, caused by a horizontal force of 1 k acting at A, considering both flexuraland axial strains in the rib. The horizontal displacement of A, resulting from the rib shorteningcaused by the real loads, is given by ∑NδN

AE∆L (28.16)

where N is the axial thrust taken from Table 28.7, δN is the axial thrust caused by a 1 k fictitioushorizontal force acting at A and equal to cos α in Table 28.1, ∆L is the length of each segment,and A is the right sectional area.

2. The horizontal displacement of A, resulting from the rib shortening caused by a 1 k horizontalforce acting at A, is given by ∑ δN

AEN∆L (28.17)

3. The necessary computations for evaluating these two deflections of A are shown in Table 28.9.

4. The horizontal displacement of A, resulting from the rib shortening caused by the real load, is∑δN

N

AE∆L =

2× 87.88× 10.798× 1230, 000

= 0.760 in (28.18)

5. That resulting from the rib shortening caused by a 1 k horizontal load at A is

∑δN

N

AE∆L =

2× 0.2300× 10.798× 1230, 000

= 0.002 in (28.19)

6. The relative value of the horizontal displacement of A, resulting from the flexural strain causedby a 1 k horizontal force at A, has previously been determined in column (3) of Table ??. Theabsolute value for this displacement is

2× 3.019× 10.798× 1728× 172830, 000

= 3.755 in . (28.20)

7. The total value for the displacement of A due to both flexural and axial strains resulting from a 1k horizontal force at A is therefore

3.755 + 0.002 = 3.757 in (28.21)

8. The reduction in HA due to rib shortening is given by

0.7603.757

= 0.20 k (28.22)



9. An additional reduction in HA will result from any drop in the temperature of the rib belowthat temperature at which the arch is erected. This occurs, of course, because some additionalshortening of the rib will result. The decrease in the length of the horizontal projection of the ribresulting from a temperature drop of 100◦F will be

δL = 0.0000065× 100× 200× 12 = 1.56 in (28.23)

10. The resulting decrease in HA will be1.563.757

= 0.42 k (28.24)

11. The total decrease in HA due to both rib shortening and temperature drop is

0.20 + 0.42 = 0.62 k (28.25)

This will have the effect of increasing the moment near the center of the span.

12. Inspection of column (6) in Table 28.8 indicates that segment 10 should be checked for possibleoverstress as the result of this increase. The increase in moment at segment 10 will be given by

∆M = ∆HA(yfor segment10) = 0.62× 34.91 = 21.6 ft · k (28.26)

13. The moment of 100 ft·k, as previously determined for segment 10, is accurate only in the first twoplaces, and hence the revised total moment for segment 10 will be

100 + 20 = 120 ft · k (28.27)

14. The thrust of 335 k, previously determined for segment 10, is accurate to three places. The changein HA due to rib shortening and temperature change must therefore be rounded to 1 k in order todetermine the corrected thrust in segment 10. This corrected thrust will be

335− 1 = 334 k (28.28)

15. The extreme fiber stress is

f =Mc

I=

120× 8.23× 121631

= 7.25 k/in2 (28.29)

16. The section at segment 10 is checked, as indicated in Table 28.8, by

N/A

FA+Mc/I

FB=

334/34.8516.01

+7.2520

= 0.60 + 0.36 = 0.96 (28.30)

and obviously the section at segment 10 is adequate.

17. No consideration has been given to the effects of a temperature rise, since such a rise can occuronly during the summer months when no snow loading is possible. Consequently, this phase of theproblem has no practical significance.

18. The method of analysis just demonstrated will apply equally well to a reinforced concrete or atimber two-hinged arch. As previously stated, however, very few reinforced concrete arches arebuilt with two hinges.

19. Of special note here is the fact that, in addition to axial strain and temperature drop, the shrinkageof the concrete in a reinforced concrete rib will also cause rib shortening. The shrinkage coefficientshould be determined for the mix to be used and applied in the analysis in the same general wayas demonstrated above for the temperature coefficient.


Draft

Chapter 29

INFLUENCE LINES (unedited)

UNEDITED

An influence line is a diagram whose ordinates are the values of somefunction of the structure (reaction, shear, moment, etc.) as a unit loadmoves across the structure.

The shape of the influence line for a function (moment, shear, reaction,etc.) can be obtained by removing the resistance of the structure to thatfunction, at the section where the influence line is desired, and applyingan internal force associated with that function at the section so as toproduce a unit deformation at the section. The deformed shape that thestructure will take represents the shape of the influence line.

Draft29–2 INFLUENCE LINES (unedited)


Draft

Chapter 30

ELEMENTS of STRUCTURALRELIABILITY

30.1 Introduction

1 Traditionally, evaluations of structural adequacy have been expressed by safety factors SF = CD , where

C is the capacity (i.e. strength) and D is the demand (i.e. load). Whereas this evaluation is quite simpleto understand, it suffers from many limitations: it 1) treats all loads equally; 2) does not differentiatebetween capacity and demands respective uncertainties; 3) is restricted to service loads; and last butnot least 4) does not allow comparison of relative reliabilities among different structures for differentperformance modes. Another major deficiency is that all parameters are assigned a single value in ananalysis which is then deterministic.

2 Another approach, a probabilistic one, extends the factor of safety concept to explicitly incorporateuncertainties in the parameters. The uncertainties are quantified through statistical analysis of existingdata or judgmentally assigned.

3 This chapter will thus develop a procedure which will enable the Engineer to perform a reliabilitybased analysis of a structure, which will ultimately yield a reliability index. This is turn is a “universal”indicator on the adequacy of a structure, and can be used as a metric to 1) assess the health of astructure, and 2) compare different structures targeted for possible remediation.

30.2 Elements of Statistics

4 Elementary statistics formulaes will be reviewed, as they are needed to properly understand structuralreliability.

5 When a set of N values xi is clustered around a particular one, then it may be useful to characterizethe set by a few numbers that are related to its moments (the sums of integer powers of the values):

Mean: estimates the value around which the data clusters.

µ =1N

N∑i=1

xi (30.1)

it is the arithmetic average of all the data points.

Expected Value: If data are not available, an expected value is assigned based on experience andjudgment, and E(x) = µx.

Both the mean and the expected values are termed first moment, and they correspond to thecentroid of a probability density distribution.

E(x) = µx

{=∫∞−∞ xf(x)dx Continuous systems

=∑N

i=1 xf(x) Discrete systems(30.2)

Draft30–2 ELEMENTS of STRUCTURAL RELIABILITY

Median: of a sorted series (xi−1 < xi < xi+1) is defined as:

xmed =

{xN+1

2N odd

12 (xN2 + xN

2 +1) N even

(30.3)

Variance: is an indication of the “width” of the cluster:

σ2 =1

N − 1ΣNi=1(xi − µ)2 (30.4)

or

σ2 =1N

ΣNi=1(xi − µ)2 (30.5)Notethat if N is less than 10, it is more appropriate to use the second equation, otherwise use the firstone.

Standard Deviation: is defined as the square root of the Variance

σ =√σ2 (30.6)

Coefficient of Variation: is the standard deviation normalized with respect to the mean:

V =σ

µ(30.7)

When insufficient data are available to accurately compute the moments, the coefficient of variationis often estimated on the basis of experience.

Covariance: Pairs of random variables may be correlated or independent. If correlated, then thelikelihood of y depends on the likelihood of x. Thus, covariance σxy measures the combined effectof how two variables vary together.

σxy =1N

N∑i=1

(xi − µx)(yi − µy) (30.8)

Correlation Coefficient: ρxy is a nondimentional measure of the degree of correlation

ρxy =σxyσxσy

(30.9)

A correlation coefficient of 1.0 or −1.0 indicates a perfect linear correlation. A positive valueindicates that the variables either increase or decrease together, a negative one indicates that onevalue increases while the other decreases. A zero value indicates that there is no linear correlationbetween the variables.

Skewness: characterizes the degree of asymmetry of a distribution around its mean. It is defined in anon-dimensional value. A positive one signifies a distribution with an asymmetric tail extendingout toward more positive x

Skew =1N

ΣNi=1

[xi − µσ

]3(30.10)

Kurtosis: is a nondimensional quantity which measures the “flatness” or “peakedness” of a distribution.It is normalized with respect to the curvature of a normal distribution. Hence a negative valuewould result from a distribution resembling a loaf of bread, while a positive one would be inducedby a sharp peak:

Kurt =1N

ΣNi=1

[xi − µσ

]4− 3 (30.11)

the −3 term makes the value zero for a normal distribution.

6 The expected value (or mean), standard deviation and coefficient of variation are interdependent:knowing any two, we can determine the third.


Draft30.3 Distributions of Random Variables 30–3

30.3 Distributions of Random Variables

7 Distribution of variables can be mathematically represented.

30.3.1 Uniform Distribution

8 Uniform distribution implies that any value between xmin and xmax is equaly likely to occur.

30.3.2 Normal Distribution

9 The general normal (or Gauss) distribution is given by, Fig. 30.1:

−3.0 −2.5 −2.0 −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0x

0.00.10.20.30.40.50.60.70.80.91.0

CD

F

−3.0

−3.0

−2.5

−2.5

−2.0

−2.0

−1.5

−1.5

−1.0

−1.0

−0.5

−0.5

0.0

0.0

0.5

0.5

1.0

1.0

1.5

1.5

2.0

2.0

2.5

2.5

3.0

3.0

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

PD

F

2σ

µ

Figure 30.1: Normalized Gauss Distribution, and Cumulative Distribution Function

φ(x) =1√2πσ

e−12 [ x−µσ ]2 (30.12)

10 A normal distribution N(µ, σ2) can be normalized by defining

y =x− µσ

(30.13)

and y would have a distribution N(0, 1):

φ(y) =1√2πe−

y2

2 (30.14)

11 The normal distribution has been found to be an excellent approximation to a large class of distribu-tions, and has some very desirable mathematical properties:



1. f(x) is symmetric with respect to the mean µ.

2. f(x) is a “bell curve” with inflection points at x = µ± σ.3. f(x) is a valid probability distribution function as:∫ ∞

−∞f(x) = 1 (30.15)

4. The probability that xmin < x < xmax is given by:

P (xmin < x < xmax) =∫ xmax

xmin

f(x)dx (30.16)

5. Cumulative distribution functions (cdf) of the normal distribution defined as:

Φ(s) =1√2πσ

∫ s

−∞e−

12 [ x−µσ ]2dx (30.17)

and is expressed in terms of the error function (erf).

6. The cdf of normalized normal distribution function is given by:

Φ(s) =1√2π

∫ s

−∞e−

x22 dx (30.18)

and is usually tabulated in books.

30.3.3 Lognormal Distribution

12 A random variable is lognormally distributed if the natural logarithm of the variable is normallydistributed.

13 It provides a reasonable shape when the coefficient of variation is large.

30.3.4 Beta Distribution

14 Beta distributions are very flexible and can assume a variety of shapes including the normal anduniform distributions as special cases.

15 On the other hand, the beta distribution requires four parameters.

16 Beta distributions are selected when a particular shape for the probability density function is desired.

30.3.5 BiNormal distribution

30.4 Reliability Index

30.4.1 Performance Function Identification

17 Designating F the capacity to demand ratio C/D (or performance function), in general F is a functionof one or more variables xi which describe the geometry, material, loads, and boundary conditions

F = F (xi) (30.19)

and thus F is in turn a random variable with its own probability distribution function, Fig. 30.2.

18 A performance function evaluation typically require a structural analysis, this may range from asimple calculation to a detailled finite element study.


Draft30.4 Reliability Index 30–5

30.4.2 Definitions

19 Reliability indices, β are used as a relative measure of the reliability or confidence in the ability ofa structure to perform its function in a satisfactory manner. In other words they are a measure of theperformance function.

20 Probabilistic methods are used to systematically evaluate uncertainties in parameters that affectstructural performance, and there is a relation between the reliability index and risk.

21 Reliability index is defined in terms of the performance function capacity C, and the applied load ordemand D. It is assumed that both C and D are random variables.

22 The safety margin is defined as Y = C −D. Failure would occur if Y < 0 Next, C and D can becombined and the result expressed logarithmically, Fig. 30.2.

X = lnC

D(30.20)

Failure would occur for negative values of X

23 The probability of failure Pf is equal to the ratio of the shaded area to the total area under thecurve in Fig. 30.2.

0.0 100.0 200.0 300.0 400.0 500.0Tensile Stress [psi]

0

20

40

60

80

100

120

140

PD

F

σ=300; µ=50σ=150; µ=30

Demand

Capacity

Failure

−1.00 −0.50 0.00 0.50 1.00 1.50 2.00ln(C/D)

0

2

4

6

8

10

pdf l

n(C

/D)

βσ

µ

ln(C/D)

Failure

Figure 30.2: Definition of Reliability Index

24 If X is assumed to follow a Normal Distribution than it has a mean value X =(ln C

D

)m

and astandard deviation σ.

25 We define the safety index (or reliability index) as the distance between mean performance value

and the limit state normalized with respect to the standard deviation

β =µln C/D

σln C/D

(30.21)

=lnµ C/D − σ2ln C/D

2√ln[1 +

(σ C/D

µ C/D

)2] (30.22)

26 The limit state is defined as a ratio of capacity to demand approaching unity (i.e. ln CD → 0).

27 Alternatively, the reliability index can be expressed in terms of the individual means of C and D, andtheir respective coefficients of variations

β =ln µC

µD√V 2C + V 2

D

(30.23)



28 For standard distributions and for β = 3.5, it can be shown that the probability of failure is Pf = 19,091

or 1.1 × 10−4. That is 1 in every 10,000 structural members designed with β = 3.5 will fail because ofeither excessive load or understrength sometime in its lifetime.

30.4.3 Mean and Standard Deviation of a Performance Function

29 In general, some of the analysis parameters are constants, and others are variables. The variables willbe characterized by a mean and a standard deviation, and a distribution function.

30 The objective is to determine the mean and standard deviation of the performance function definedin terms of C/D.

31 Those two parameters, in turn, will later be required to compute the reliability index.

30.4.3.1 Direct Integration

32 Given a function random variable x, the mean value of the function is obtained by integrating thefunction over the probability distribution function of the random variable

µ[F (x)] =∫ ∞

−∞g(x)f(x)dx (30.24)

33 For more than one variable,

µ[F (x)] =∫ ∞

−∞

∫ ∞

−∞· · ·∫ ∞

−∞F (x1, x2, · · · , xn)F (x1, x2, · · · , xn)dx1dx2 · · · dxn (30.25)

34 Note that in practice, the function F (x) is very rarely available for practical problems, and hence thismethod is seldom used.

30.4.3.2 Monte Carlo Simulation

35 The performance function is evaluated for many possible values of the random variables.

36 Assuming that all variables have a normal distribution, then this is done through the following algo-rithm

1. initialize random number generators

2. Perform n analysis, for each one:

(a) For each variable, determine a random number for the given distribution

(b) Transform the random number

(c) Analyse

(d) Determine the performance function, and store the results

3. From all the analyses, determine the mean and the standard deviation, compute the reliabilityindex.

4. Count the number of analyses, nf which performance function indicate failure, the likelihood ofstructural failure will be p(f) = nf/n.

37 A sample program (all subroutines are taken from (Press, Flannery, Teukolvsky and Vetterling 1988)which generates n normally distributed data points, and then analyze the results, determines mean andstandard deviation, and sort them (for histogram plotting), is shown below:



program nice

parameter(ns=100000)

real x(ns), mean, sd

write(*,*)’enter mean, standard-deviation and n ’

read(*,*)mean,sd,n

c initialize

tmp=gasdev(-1)

do i=1,n

c obtain gausian distribution and adjust for mean and standard deviation

x(i)=gasdev(1)*sd+mean

end do

c compute mean and standard deviation

call moment(x,n,ave,adev,sdev,var,skew,curt)

c sort and print results

call sort(n,x)

write(*,*)(x(i),i=1,n)

write(*,*)’ average standard deviation’,ave,sdev

stop

end

c-----------------------------------------------------------------------------

function gasdev(idum)

c*** Function to produce a normally distributed variable with zero

c mean and unit ariance.

data iset/0/

if (iset.eq.0) then

1 v1=2.*ran1(idum)-1.

v2=2.*ran1(idum)-1.

r=v1**2+v2**2

if(r.ge.1..or.r.eq.0.)go to 1

fac=sqrt(-2.*log(r)/r)

gset=v1*fac

gasdev=v2*fac

iset=1

else

gasdev=gset

iset=0

endif

return

end

c-----------------------------------------------------------------------------

function ran1(idum)

c *** Returns a uniform random variable between 0.0 and 1.0,

c set idum to any negative value to initialize, or reinitialize the sequence

dimension r(97)

parameter (m1=259200,ia1=7141,ic1=54773,rm1=3.8580247e-6)

parameter (m2=134456,ia2=8121,ic2=28411,rm2=7.4373773e-6)

parameter (m3=243000,ia3=4561,ic3=51349)

data iff /0/

if (idum.lt.0.or.iff.eq.0) then

iff=1

ix1=mod(ic1-idum,m1)

ix1=mod(ia1*ix1+ic1,m1)

ix2=mod(ix1,m2)


ix3=mod(ix1,m3)

do 11 j=1,97



r(j)=(float(ix1)+float(ix2)*rm2)*rm1

11 continue

idum=1

endif




j=1+(97*ix3)/m3

if(j.gt.97.or.j.lt.1)pause

ran1=r(j)

r(j)=(float(ix1)+float(ix2)*rm2)*rm1

return

end

c-----------------------------------------------------------------------------



subroutine moment(data,n,ave,adev,sdev,var,skew,curt)

c given array of data of length N, returns the mean AVE, average

c deviation ADEV, standard deviation SDEV, variance VAR, skewness SKEW,

c and kurtosis CURT

dimension data(n)

if(n.le.1)pause ’n must be at least 2’

s=0.

do 11 j=1,n

s=s+data(j)

11 continue

ave=s/n

adev=0.

var=0.

skew=0.

curt=0.

do 12 j=1,n

s=data(j)-ave

adev=adev+abs(s)

p=s*s

var=var+p

p=p*s

skew=skew+p

p=p*s

curt=curt+p

12 continue

adev=adev/n

var=var/(n-1)

sdev=sqrt(var)

if(var.ne.0.)then

skew=skew/(n*sdev**3)

curt=curt/(n*var**2)-3.

else

pause ’no skew or kurtosis when zero variance’

endif

return

end

c------------------------------------------------------------------------------

subroutine sort(n,ra)

dimension ra(n)

l=n/2+1

ir=n

10 continue

if(l.gt.1)then

l=l-1

rra=ra(l)

else

rra=ra(ir)

ra(ir)=ra(1)

ir=ir-1

if(ir.eq.1)then

ra(1)=rra

return

endif

endif

i=l

j=l+l

20 if(j.le.ir)then

if(j.lt.ir)then

if(ra(j).lt.ra(j+1))j=j+1

endif

if(rra.lt.ra(j))then

ra(i)=ra(j)

i=j

j=j+j

else

j=ir+1

endif

go to 20

endif

ra(i)=rra

go to 10

end



38 As an example, using the above program, and assuming a normal distribution with a mean of 100.and a standard deviation of 20, results shown in Table 30.1 are obtained.

n µ σ xmin xmax10 100.33 7.17 91.05 111.29100 100.49 9.73 77.01 120.92

1,000 100.123 9.73 72.54 131.4310,000 99.964 9.88 66.20 134.31100,000 100.029 9.97 53.65 145.39

Table 30.1: Tabulated Results of a Simple Monte-Carlo Simulation.

39 Note that for monte carlo simulations to be effective, hundreds (and occasionally thousands) of anal-yses must be performed.

30.4.3.3 Point Estimate Method

40 As noted above, even for the simplest problem a successful Monte-Carlo simulation can be computa-tionally very expensive. This is even more though when the evaluation of the performance function canonly be achieved through a finite element analysis.

41 Whereas in the Monte Carlo simulations numerous random analyses had to be performed, in thismethod we select only few points such that they match the mean, standard deviation, and coefficient ofskewness of the distribution1, (Harr 1987, Wolff 1985).

42 This is accomplished by limiting ourselves to all possible combinations of µi ± σi.43 For each analysis we determine SFF, as well as its logarithm.

44 Mean and standard deviation of the logarithmic values are then determined from the 2n analyses,and then β is the ratio of the mean to the standard deviation.

30.4.3.4 Taylor’s Series-Finite Difference Estimation

45 In the previous method, we have cut down the number of deterministic analyses to 2n, in the followingmethod, we reduce it even further to 2n+ 1, (Anonymous 1992, Anonymous 1993, Bryant, Brokaw andMlakar 1993).

46 This simplified approach starts with the first order Taylor series expansion of Eq. 30.19 about themean and limited to linear terms, (Benjamin and Cornell 1970).

µF = F (µi) (30.26)

where µi is the mean for all random variables.

47 For independent random variables, the variance can be approximated by

V ar(F ) = σ2F =∑(

∂F

∂xiσi

)2(30.27-a)

∂F

∂xi≈ F+

i − F−i

2σi(30.27-b)

1Such a simplification is behind the development of the Whitney stress block in the ACI code; a complex stressdistribution is replaced by a simpler one (rectangular), such that both have the same resultant force and location.



F+i = F (µ1, · · · , µi + σi, · · · , µn) (30.27-c)F−i = F (µ1, · · · , µi − σi, · · · , µn) (30.27-d)

where σi are the standard deviations of the variables. Hence,

σF =∑(

F+i − F−

i

2

)(30.28)

Finally, the reliability index is given by

β =lnµFσF

(30.29)

48 The procedure can be summarized as follows:

1. Perform an initial analysis in which all variables are set equal to their mean value. This analysisprovides the mean µ.

2. Perform 2n analysis, in which all variables are set equal to their mean values, except variable i,which assumes a value equal to µi + σi, and then µi − σi.

3. For each pair of analysis in which variable xi is modified, determine

(a) The standard deviation, which will provide an indication of the sensitivity of the results tovariation of this particular variable.

(b)(F+i−F−

i

2

)4. The standard deviation is then determined by simply adding all the

(F+i−F−

i

2

)terms. from the 2n

analyses. by the mean.

5. Having determined the mean and standard deviation of C/D, the corresponding values for thelogarithm of C/D, µlnC/D and σlnC/D, and β are determined from Eq. 30.22.

30.4.4 Overall System Reliability

49 Reliability indices for a number of components, or a number of modes of performance may be usedto estimate the overall reliability of a structure.

50 We consider two extreme cases:

Series System: Such a system will perform poorly if any one component perform unsatisfactorily. Ifa system has n components in series, the probability of unsatisafcatorily performance of the ithcomponent is pi and its reliability Ri = 1 − pi, then the reliability of the system, or probabilitythat all components will perform satisfactorily is

R = R1R2 · · ·Ri · · ·Rn= (1− pi)(1− p2) · · · (1− pi) · · · (1 − pn) (30.30-a)

Parallel Systems: Such a system will only perform unsatisfactorily if all component perform unsatis-factorily. Hence the reliability is

R = 1− p1p2 · · · pi · · · pn (30.31)

30.4.5 Target Reliability Factors

51 Reliability indices are a relative measure of the current condition and provide a qualitative estimateof the structural performance.

52 Structures with relatively high reliable indices will be expected to perform well. If the value is toolow, then the structure may be classified as a hazard.

53 Target values for β are shown in Table 30.2, and in Fig. 30.3



0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5β

100

101

102

103

104

105

106

107

1/(P

roba

bilit

y of

Fai

lure

)

Probability of Failure in terms of β

Go

od

Haz

ard

ou

s

Un

sati

sfac

tory

Po

or

Bel

ow

Ave

rag

e

Ab

ove

Ave

rag

e

Figure 30.3: Probability of Failure in terms of β



Expected Performance β FailuresHigh 5 3/10 millionGood 4 3/100,000Above Average 3 1/1,000Below Average 2.5 6/1,000Poor 2.0 2.3/100Unsatisfactory 1.5 7/100Hazardous 1.0 16/100

Table 30.2: Selected β values for Steel and Concrete Structures

30.5 Reliability Analysis

The steps involved in a reliability analysis are:

1. Model Develop a model which describes the performance mode and limit state to be evaluated.

2. Limit state determine the limit states in terms of ratios of capacity to demand.

3. Parameters Identify material properties and structure geometric parameters which enter into theanalysis/ Separate those which may be considered constants from those which should be treatedas random variables.

4. Variables Assign means, standard deviation, and coefficients of correlations to random variablesbased on available or obtainable information. These may be derived from distributions, calculateddirectly from data, or assigned judgmentally.

5. Performance function Formulate the performance function C/D, in terms of constants and variablesthat describe the performance mode of interest.

6. Means and standard deviations of performance functions Determine the means and standard de-viations of the performance function (C/D) for the performance modes of interest.

7. Calculate reliability index β Calculate β using any of the methods described above.


Draft

Bibliography

318, A. C.: n.d., Building Code Requirements for Reinforced Concrete, (ACI 318-83), American ConcreteInstitute.

Anonymous: 1992, Reliability assessment of navigation structures, ETL 1110-2-532, Department of theArmy, US Army Corps of Engineers, Washington, D.C.

Anonymous: 1993, Reliability assessment of navigation structures; stability of existing gravity structures,ETL 1110-2-321, Department of the Army, US Army Corps of Engineers, Washington, D.C.

Arbabi, F.: 1991, Structural Analysis and Behavior, McGraw-Hill, Inc.

Benjamin, J. and Cornell, C.: 1970, Probability, Statistics, and Decision for Civil Engineers, McGrawHill, New York.

Billington, D.: 1979, Robert Maillart’s Bridges; The art of Engineering, Princeton University Press.

Billington, D. and Mark, R.: 1983, Structural studies, Technical report, Department of Civil Engineering,Princeton University.

Bryant, L., Brokaw, J. and Mlakar, P.: 1993, Reliability modeling of concrete overstressing, Technicalreport, JAYCOR, Vicksburg Mississippi. Report Submitted to U.S. Army Engineer WaterwaysExperiment Station.

Chajes, A.: 1983, Prentice-Hall.

Gerstle, K.: 1974, Basic Structural Analysis, XX. Out of Print.

Harr, M.: 1987, Reliability Based Design in Civil Engineering, McGraw Hill, New York.

Kinney, J.: 1957, Indeterminate Structural Analysis, Addison-Wesley. Out of Print.

Lin, T. and Stotesbury, S.: 1981, Structural Concepts and Systems for Architects and Engineers, JohnWiley.

of Steel COnstruction, A. I.: 1986, Manual of Steel Construction; Load and Resistance Factor Design,American Institute of Steel Construction.

Penvenuto, E.: 1991, An Introduction to the History of Structural Mechanics, Springer-Verlag.

Press, W., Flannery, B., Teukolvsky, S. and Vetterling, W.: 1988, Numerical Recipies in Fortran, TheArt of Scientific Computing, Cambridge University Press.

Salmon, C. and Johnson, J.: 1990, Steel Structures; Design and Behavior, Harper Colins. Third Edition.

Schueller, W.: 1996, The design of Building Structures, Prentice Hall.

UBC: 1995, Uniform building code, Technical report, International COnference of Building Officials.

White, R., Gergely, P. and Sexmith, R.: 1976, Structural Engineering, John Wiley & Sons. Out of Print.

Wolff, T.: 1985, Analysis and Design of Embankment Dam Slopes: A Probabilistic Approach, PhD thesis,Purdue University.

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